Target Audience: Calculus students beginning series theory.
Prerequisites: Basic algebra and function notation. Two parts of the page go beyond a first calculus course; the reading note below says which.
The page is built to be read twice. On a first pass, take Sections 1–6, then Section 7 as far as its two worked examples, then Sections 8–12. That is a complete first course: what a series is, what it means for one to converge, and the two families whose sums you can actually compute.
On a second pass, add the rest of Section 7 — subsequences, Bolzano–Weierstrass and Cauchy sequences, which is where the theory really lives — and Section 13, which turns a series into a function and opens the door to the advanced tutorial. Nothing in the first pass depends on them; everything in the intermediate tutorial ultimately does.
Understanding the difference between a sequence and a series is the foundation of series theory.
A sequence is a function whose domain is the natural numbers: to each \(n = 1, 2, 3, \ldots\) it assigns a real number, written \(a_n\) rather than \(a(n)\) and called the \(n\)-th term. Listing the values in order gives the familiar picture of an ordered list:
Example: The sequence \(\left(\frac{1}{n}\right)\) is \(\left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\right)\)
The definition asks for no formula and no pattern. \(a_n = \frac{1}{n}\) is a sequence, but so is the sequence of decimal digits of \(\pi\), and so is one produced by tossing a coin forever. What a sequence has that a set does not is order — inherited from the ordering of \(\mathbb{N}\) — which is why \((1, 2, 3, \ldots)\) and \((3, 1, 2, \ldots)\) are different sequences even though \(\{1, 2, 3\}\) and \(\{3, 1, 2\}\) are the same set.
You will also meet the braces form \(\{a_n\}\). It is common, but braces are set notation, and a set has no order; round brackets are the notation that says what a sequence is.
A series is the sum of the terms of a sequence:
Adding infinitely many numbers is not an operation we have yet, so for now read this as the intended sum. What the expression actually means is made precise in Section 8, through partial sums.
Example: The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\) (the harmonic series)
| Concept | Sequence | Series |
|---|---|---|
| Nature | An ordered list of numbers | The sum of the terms of a sequence |
| Notation | \((a_n)\) or \(\{a_n\}\) | \(\sum a_n\) |
| Question Asked | What do the terms approach? | What does the sum approach? |
Sigma notation provides a compact way to write sums. The Greek letter \(\Sigma\) (sigma) stands for "sum."
You can change the index variable without changing the sum's value:
Both sums equal \(a_1 + a_2 + a_3 + a_4 + a_5\).
An arithmetic sequence has a constant difference \(d\) between consecutive terms.
where \(a_1\) is the first term and \(d\) is the common difference.
We want a simple formula for the sum of the first \(n\) terms of an arithmetic sequence \((a_n)\) with first term \(a_1\) and common difference \(d\).
The sum of the first \(n\) terms is
Using the general term \(a_n = a_1 + (n-1)d\), this is the same as
Write the sum once in the usual order and once in reverse order, lining terms up vertically:
Each column now contains one term from the top line and the matching term from the bottom line.
Add the two displayed equations side by side. Each pair of terms (first with last, second with second-last, etc.) gives the same sum:
There are exactly \(n\) such identical pairs, so
Divide both sides by 2 to isolate \(S_n\):
Sometimes it is more convenient to write the formula in terms of the first and last terms. Since \(a_n = a_1 + (n-1)d\), we can replace \(2a_1 + (n-1)d\) by \(a_1 + a_n\):
You can read this as: sum = number of terms × average of first and last term.
The sum \(1 + 2 + 3 + \dots + n\) is a special case where \(a_1 = 1\), \(d = 1\), and \(a_n = n\).
Using the "average of first and last" version,
so
On the interactive slider below, this is exactly the value being computed for each choice of \(n\).
\(S_n = \frac{n(n+1)}{2}\) = 55
Drag the slider to see how the sum grows with n.
A geometric sequence has a constant ratio \(r\) between consecutive terms.
where \(a_1\) is the first term and \(r\) is the common ratio.
Let \(S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}\). Multiply by \(r\):
Subtracting: \(S_n - rS_n = a_1 - a_1 r^n\), so:
With \(|r| \lt 1\), the series converges to 2.00
The chart shows partial sums \(S_n\) for \(n = 1\) to 20 with \(a_1 = 1\).
The concept of a limit captures the idea that a sequence "approaches" a value.
We say \(\lim_{n \to \infty} a_n = L\) if the terms \(a_n\) get arbitrarily close to \(L\) as \(n\) gets large.
We write \(\lim_{n \to \infty} a_n = L\) if:
"For every positive tolerance \(\epsilon\), there exists a threshold \(N\) such that all terms beyond \(N\) are within \(\epsilon\) of \(L\)."
Note that the definition says "\(a\) limit", yet we immediately write "the limit". That is licensed by a theorem, not a convention — a sequence cannot converge to two different values, proved on the Uniqueness of Limits page.
Two results get used constantly from here on, usually without comment. Both are proved on their own pages; the statements are what you need to keep in view.
If \(a_n \to A\) and \(b_n \to B\), then
and \(a_n / b_n \to A / B\) provided \(B \neq 0\). This is what makes the algebra of every later section legitimate. Proved on the Limit Laws page.
If \(a_n \leq b_n \leq c_n\) for all sufficiently large \(n\), and \(a_n \to L\) and \(c_n \to L\), then
The sequence in the middle has nowhere else to go. Proved on the Squeeze Theorem page; the geometric series section leans on it to show \(r^n \to 0\).
The terms \(1, \frac{1}{2}, \frac{1}{3}, \ldots\) approach 0.
This famous limit defines Euler's number \(e \approx 2.718\).
The informal claim that $\frac{1}{n}$ "approaches 0" can be made rigorous directly from the $\epsilon$-$N$ definition above.
Goal: For each $\epsilon \gt 0$, produce an $N$ such that $n \gt N \Rightarrow \left|\frac{1}{n} - 0\right| \lt \epsilon$ — matching the $\epsilon$-$N$ definition above, which uses strict $n \gt N$.
Step 1 — Simplify the inequality.
For positive $n$, the condition $\frac{1}{n} \lt \epsilon$ is equivalent to:
Step 2 — Choose $N$.
Let $N = \left\lceil \frac{1}{\epsilon} \right\rceil$ (the ceiling function — the smallest integer $\geq 1/\epsilon$). This is a positive integer for every $\epsilon \gt 0$, as the definition requires.
Step 3 — Verify.
Suppose $n \gt N$. Then $n \gt N \geq \frac{1}{\epsilon}$, so $n \gt \frac{1}{\epsilon}$, and Step 1 gives $\frac{1}{n} \lt \epsilon$ — exactly what was required. $\blacksquare$
The strict inequality is doing real work here. Note that $N$ itself only satisfies $N \geq 1/\epsilon$, not $N \gt 1/\epsilon$; the strictness comes from $n \gt N$. Relax the goal to $n \geq N$ and the argument breaks whenever $1/\epsilon$ is an integer, because then $\lceil 1/\epsilon \rceil = 1/\epsilon$ exactly and $n = N$ gives $\frac{1}{n} = \epsilon$, not $\lt \epsilon$. Concretely, with $\epsilon = \tfrac{1}{2}$ we get $N = 2$, and $\frac{1}{2} \lt \frac{1}{2}$ is false. This is why the $\epsilon$-$N$ definition above is stated with strict $n \gt N$ — and why it pays to match your proof to the definition exactly.
Intuition: The smaller $\epsilon$ is, the further out in the sequence you must go before the terms enter the band. As $\epsilon \to 0$, $N$ grows without bound, which correctly reflects that the sequence never actually reaches 0 — it only approaches it.
Move the slider to choose a tolerance.
Grey points lie outside the band \(|a_n - L| \lt \epsilon\); green points are inside. The definition asks for a threshold \(N\) after which every point is green. Shrink \(\epsilon\) and watch \(N\) move right — that is the "for every \(\epsilon\), there exists \(N\)" of the definition, one \(\epsilon\) at a time.
A sequence with no limit diverges. That one word covers two quite different behaviours, and it pays to name them, because both reappear for series.
We write \(a_n \to +\infty\) if the terms eventually exceed any bound:
Similarly \(a_n \to -\infty\). Example: \(a_n = n^2\). Note the wording: such a sequence still diverges. Infinity is not a real number, so it is never a limit in the sense of the definition above — the arrow describes how the sequence fails to have one.
A sequence can stay bounded and still have no limit:
It keeps returning to two different values, so no single \(L\) can capture its tail. Every convergent sequence is bounded — its tail sits within \(\epsilon\) of \(L\) and only finitely many terms lie outside — but the converse fails, as this example shows. The gap between bounded and convergent is exactly what the monotone and bounded section is about.
A recursive sequence (or recurrence relation) defines each term from the ones before it, rather than by a formula in \(n\). You are given a starting value and a rule.
The first few terms, to four places:
They climb, each step smaller than the last, and they seem to be heading for \(2\). Two things are being claimed there — that the sequence converges, and that its limit is \(2\) — and it matters that they are different claims, settled by different means.
Nothing so far proves either. Six terms are six terms. The tool that decides convergence without computing the limit is the subject of the next section, and this sequence is its first worked example there.
It is tempting to reason: "if \(a_n \to L\), then \(a_{n+1} \to L\) as well, so \(L = \sqrt{2 + L}\), giving \(L^2 - L - 2 = 0\) and \(L = 2\)." The algebra is right, and it will be used — but only after convergence is established. The argument begins with "if \(a_n \to L\)": it identifies what the limit must be if there is one, and says nothing about whether there is.
The gap is real. Apply the same reasoning to \(a_1 = 1\), \(a_{n+1} = 2a_n\): "if \(a_n \to L\) then \(L = 2L\), so \(L = 0\)." But the sequence is \(1, 2, 4, 8, \ldots\), which diverges to infinity. The fixed-point equation returned an answer for a sequence that has no limit at all.
Perhaps the most famous recursive sequence, and one whose rule reaches back two terms rather than one:
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| \(F_n\) | 0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 |
Fun fact: The ratio \(\frac{F_{n+1}}{F_n}\) approaches the golden ratio \(\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618\). The sequence returns in the advanced tutorial, where a power series built from it — its generating function — yields a closed formula for \(F_n\).
These properties let us determine whether a sequence converges without finding the limit explicitly — which matters enormously for series, where the sum is usually unknown and often unknowable in closed form.
This section is the engine room. Everything later in this tutorial — and every convergence test in the intermediate one — ultimately rests on a single property of the real numbers, developed here.
Everything below follows from one property, and it is an axiom — not a theorem. It is the defining feature of the real numbers.
Every nonempty set of reals that is bounded above has a least upper bound (a supremum).
The dual statement follows immediately: every nonempty set bounded below has a greatest lower bound (an infimum).
Note it is stated for sets, not sequences. That generality is what makes it usable as a foundation.
This is precisely what \(\mathbb{R}\) has and \(\mathbb{Q}\) lacks. The set \(\{q \in \mathbb{Q} : q^2 \lt 2\}\) is nonempty and bounded above within \(\mathbb{Q}\), yet has no rational least upper bound — the number that ought to be its supremum, \(\sqrt{2}\), simply is not there. The rationals have a hole exactly where the answer should be.
That is not a curiosity. It is the whole reason convergence works in \(\mathbb{R}\): a sequence of rationals can march steadily toward a target that does not exist in \(\mathbb{Q}\). Completeness is the guarantee that in \(\mathbb{R}\), the target is always there to be reached.
Every bounded monotone sequence converges.
The proof is short and worth seeing: because the supremum \(L\) is the least upper bound, \(L - \epsilon\) cannot be an upper bound, so some term already exceeds it — and monotonicity then drags the entire tail into \((L - \epsilon, L]\). Written out, together with Bolzano–Weierstrass, on the Monotone Convergence and Bolzano–Weierstrass proof page.
Return to \(a_1 = \sqrt{2}\), \(a_{n+1} = \sqrt{2 + a_n}\). The theorem needs two facts, and both go by induction.
Bounded above by 2. \(a_1 = \sqrt{2} \lt 2\). If \(a_n \lt 2\), then \(a_{n+1} = \sqrt{2 + a_n} \lt \sqrt{2 + 2} = 2\). So every term is below \(2\).
Increasing. All terms are positive, so \(a_{n+1} \gt a_n\) is equivalent to \(a_{n+1}^2 \gt a_n^2\), that is \(2 + a_n \gt a_n^2\), which factors as \((2 - a_n)(1 + a_n) \gt 0\). Both factors are positive because \(0 \lt a_n \lt 2\). So each term exceeds the last.
Conclusion. Bounded and monotone, so by the Monotone Convergence Theorem the sequence converges, to some \(L\) with \(\sqrt{2} \leq L \leq 2\). Only now is the fixed-point algebra legitimate: \(a_{n+1} \to L\) as well, and square roots preserve limits, so \(L = \sqrt{2 + L}\). Then \(L^2 - L - 2 = (L - 2)(L + 1) = 0\), and since \(L \gt 0\), \(L = 2\).
Notice the division of labour. The Monotone Convergence Theorem proved that the limit exists; the recurrence then told us what it is. Neither step could do the other's job.
The sequence \(a_n = \left(1 + \frac{1}{n}\right)^n\) also converges by this theorem, and here the pay-off is larger, because its limit — \(e\) — has no simpler description. Two facts again.
Increasing. A little algebra rewrites the ratio of consecutive terms as
the inequality being Bernoulli's inequality (stated in Section 11) in the form \((1 + x)^{n+1} \geq 1 + (n+1)x\) with \(x = -\frac{1}{(n+1)^2}\).
Bounded above by 3. Expand by the binomial theorem. The \(k\)-th term is \(\binom{n}{k} \frac{1}{n^k} \leq \frac{1}{k!}\), and \(\frac{1}{k!} \leq \frac{1}{2^{k - 1}}\) for \(k \geq 1\), so
The last step is the finite geometric formula of Section 4. Bounded and increasing, so the limit exists; we call it \(e\). Notice what we did not need: any way of computing \(e\). Boundedness and monotonicity alone settled the question.
Choose indices \(n_1 \lt n_2 \lt n_3 \lt \cdots\) — strictly increasing, so the terms are picked in their original order and none is picked twice. The sequence \((a_{n_k})_{k=1}^{\infty}\) of the chosen terms is a subsequence of \((a_n)\).
Example: from \(a_n = (-1)^n\), the even-indexed terms \(a_2, a_4, a_6, \ldots\) form the subsequence \(1, 1, 1, \ldots\) and the odd-indexed terms form \(-1, -1, -1, \ldots\). Each subsequence converges; the sequence itself does not.
One fact to carry forward: if \(a_n \to L\), then every subsequence also converges to \(L\), because a tail of the subsequence lives inside a tail of the original. The intermediate tutorial uses exactly this to show that bracketing a convergent series cannot change its sum.
The Monotone Convergence Theorem needs its sequence to be monotone. Most sequences are not. Bolzano–Weierstrass rescues what it can from an arbitrary bounded sequence.
Every bounded real sequence has a convergent subsequence.
The proof runs through a small combinatorial gem, the peak lemma, which says every sequence has a monotone subsequence. Call an index \(m\) a peak if \(a_m \geq a_n\) for every later \(n\). There are two cases, and both are kind to us:
Either way we obtain a monotone subsequence. It inherits boundedness from the original sequence, so the Monotone Convergence Theorem finishes the job. The full argument is on the Monotone Convergence and Bolzano–Weierstrass proof page.
Bolzano–Weierstrass promises a convergent subsequence — never that the sequence itself converges. Take \(a_n = (-1)^n\left(1 + \frac{1}{n}\right)\): it is bounded and thoroughly divergent, yet has two convergent subsequences, approaching \(+1\) and \(-1\). Boundedness buys you an accumulation point, never a limit.
Both results so far need something extra — monotonicity, or the willingness to discard most of the sequence. The Cauchy condition needs neither, and it never mentions the limit at all.
A sequence whose terms eventually get arbitrarily close to one another:
Compare the \(\epsilon\)-\(N\) definition in Section 5, which measures every term against \(L\). This one measures terms against each other — so it can be checked without knowing \(L\).
One direction is free: every convergent sequence is Cauchy, since if two terms are each within \(\epsilon/2\) of \(L\) they are within \(\epsilon\) of each other. The substantial direction is the converse — and it is exactly what completeness provides.
In \(\mathbb{R}\), every Cauchy sequence converges.
Again this is what \(\mathbb{Q}\) lacks: \(1,\, 1.4,\, 1.41,\, 1.414, \ldots\) is Cauchy in \(\mathbb{Q}\) but converges to nothing there. Proved on the Cauchy Sequences and Completeness page.
Why this matters for series: it lets you prove a series converges without producing its sum. That is not a convenience but a necessity — for most convergent series, no closed form for the sum exists. Section 8 turns this into a usable test for series.
An infinite series is defined as the limit of its partial sums.
The \(N\)-th partial sum of a series is:
The infinite series is defined as:
If this limit exists and is finite, we say the series converges. Otherwise, it diverges.
As with sequences, "diverges" covers two behaviours. The partial sums may run off to infinity, as those of the harmonic series do; or they may oscillate without settling, as for \(1 - 1 + 1 - 1 + \cdots\), whose partial sums are \(1, 0, 1, 0, \ldots\). Neither series has a sum. The second kind is the one that catches people out, because the terms need not grow at all.
We cannot literally add infinitely many numbers. Instead, we:
It is worth stating the consequence plainly: there is no separate notion of convergence for series. Every statement about a series is a statement about its sequence of partial sums. That is what lets the sequence machinery of Section 7 say anything about series at all.
The finite rules of sigma notation survive the passage to the limit, and the Limit Laws of the limits section are exactly what carries them across.
If \(\sum a_n = A\) and \(\sum b_n = B\) both converge, then
Why. The \(N\)-th partial sum of \(\sum (a_n + b_n)\) is \(S_N + T_N\), a finite sum split by the rules of sigma notation. As \(N \to \infty\), \(S_N \to A\) and \(T_N \to B\), so the Limit Laws give \(S_N + T_N \to A + B\). The other identities go the same way.
The rule runs in one direction only. \(\sum (a_n + b_n)\) can converge while \(\sum a_n\) and \(\sum b_n\) both diverge — take \(a_n = 1\) and \(b_n = -1\), whose sum is the zero series. And there is no rule at all for products: \(\sum a_n b_n\) is not \(\bigl(\sum a_n\bigr)\bigl(\sum b_n\bigr)\). Multiplying series correctly is a topic of its own, taken up in the advanced tutorial.
When every term satisfies \(a_n \geq 0\), each partial sum is at least the one before it: \(S_{N+1} = S_N + a_{N+1} \geq S_N\). The partial sums form an increasing sequence — and increasing sequences are exactly what the Monotone Convergence Theorem handles.
If \(a_n \geq 0\) for all \(n\), then \(\sum a_n\) converges if and only if its partial sums are bounded above. When they are not, \(S_N \to +\infty\).
Proof. \((S_N)\) is increasing. If it is bounded above, the Monotone Convergence Theorem says it converges, to its supremum. If it is not bounded above, then being increasing it exceeds every \(M\) from some point on and stays there — which is the definition of \(S_N \to +\infty\). \(\blacksquare\)
Note what the criterion rules out: a series of non-negative terms can never oscillate. It either settles or runs off to infinity. Oscillation needs terms of both signs.
This is the single most important consequence of the completeness material for series. Every comparison-style test in the intermediate tutorial — Direct Comparison, Limit Comparison, the Integral Test, Cauchy Condensation — is this criterion plus one inequality: trap the partial sums below something finite, and convergence follows without ever naming the sum.
Applying the Cauchy condition to the partial sums \(S_N\), rather than to the terms, gives the sharpest general statement about series convergence available.
\(\sum a_n\) converges if and only if for every \(\epsilon \gt 0\) there is an \(N\) such that
In words: every sufficiently late finite stretch of the series sums to something small. It is an "if and only if" only because \(\mathbb{R}\) is complete.
Two things fall out of this immediately, and both are worth noticing.
Take the shortest possible stretch, \(n = m + 1\). The sum collapses to the single term \(|a_n| \lt \epsilon\) — which says exactly that \(a_n \to 0\). The Divergence Test of the next section is not a separate idea; it is the Cauchy Criterion applied to stretches of length one, which is precisely why it is so weak.
There is an even shorter route that never mentions Cauchy. If \(\sum a_n = S\), then \(a_n = S_n - S_{n - 1}\), and both partial sums tend to \(S\); by the Limit Laws, \(a_n \to S - S = 0\).
Take the stretch running from \(m + 1\) to \(2m\) — that is, apply the criterion with \(n = 2m\). It holds \(m\) terms, and the smallest of them is \(\frac{1}{2m}\), so
No matter how far out you go, there is always a later stretch summing to at least \(\frac{1}{2}\). The criterion fails, so \(\sum \frac{1}{n}\) diverges — even though its terms go to zero.
Visualizing sequences and their partial sums helps build intuition for convergence vs. divergence.
Geometric series with r = 1/2: converges to 2.
The Divergence Test (also called the Term Test or n-th Term Test) is often the first convergence check to apply.
If \(\lim_{n \to \infty} a_n \neq 0\), then \(\sum_{n=1}^{\infty} a_n\) diverges.
Equivalently: For a series to have any chance of converging, its terms must approach zero.
The converse is FALSE! If \(\lim_{n \to \infty} a_n = 0\), the series may still diverge.
Counterexample: The harmonic series \(\sum \frac{1}{n}\) has terms going to 0, but the series diverges!
Bottom line: The Divergence Test can only prove divergence, never convergence. If the terms go to zero, you need other tests.
Before meeting any more tests, it is worth naming a principle they all share. Changing, adding, or removing finitely many terms of a series cannot change whether it converges — only what it converges to. Those terms contribute a finite amount, and adding a finite amount to a limit leaves its existence untouched.
So every convergence test is really a statement about the tail of a series. A hypothesis like "the terms are decreasing" or \(a_n \leq b_n\) only has to hold from some index \(N\) onward, never from \(n = 1\). A series is free to misbehave for its first thousand terms and still converge.
Why this is worth carrying forward: as the tests accumulate in the intermediate tutorial, some state their hypotheses over "all \(n\)" for readability while others say "sufficiently large \(n\)" explicitly. The two mean the same thing, and you should read every such hypothesis in the second sense — that is what is actually required, and it is what lets you apply a test to a series whose opening terms do not cooperate.
Certain series appear frequently and serve as benchmarks for comparison.
The geometric series converges if and only if \(|r| \lt 1\). This is one of the few series where we have an explicit formula for the sum.
Strictly, both halves of that "if and only if" assume \(a \neq 0\). If \(a = 0\) every term vanishes, the series converges to \(0\) for every \(r\), and the equivalence fails. A trivial case, but it is the reason the hypothesis is there.
The step that actually needs justifying is \(r^n \to 0\) when \(|r| \lt 1\). See the Geometric Series and Harmonic Series proof page for the full argument.
That last step is usually waved through, but it deserves a moment — the whole formula depends on it, and it has a clean one-line proof.
Proved by induction: multiplying through by \(1 + x \geq 0\) preserves the inequality. Full argument on the Bernoulli's Inequality page.
Applying it. Suppose \(0 \lt |r| \lt 1\). Then \(\frac{1}{|r|} \gt 1\), so we may write \(\frac{1}{|r|} = 1 + x\) for some \(x \gt 0\). Bernoulli gives \(\left(\frac{1}{|r|}\right)^n \geq 1 + nx\), and taking reciprocals flips it:
The right-hand side goes to zero because \(nx\) grows without bound, and \(0 \leq |r|^n\) on the other side, so the Squeeze Theorem forces \(|r|^n \to 0\). (The case \(r = 0\) is immediate.) So \(r^n \to 0\), and the geometric series formula stands.
The formula above starts at \(n = 0\), which is a convention rather than a mathematical requirement. Real problems rarely start there, so it is worth having the general version in your back pocket:
Setting \(k = 0\) recovers the familiar \(\frac{a}{1 - r}\). So the \(n = 0\) start is not special — it is simply the index choice that makes "the first term" and \(a\) coincide, which is why textbooks default to it.
Worked example. Take \(a = 3\), \(r = \frac{1}{2}\), starting at \(n = 2\):
Sanity check. The same series from \(n = 0\) sums to \(\frac{3}{1 - 1/2} = 6\). Starting at \(n = 2\) merely discards the first two terms, \(3\) and \(\frac{3}{2}\), so the tail must be \(6 - 3 - \frac{3}{2} = \frac{3}{2}\). It agrees.
What to remember: the numerator is always the first term you actually write down, \(ar^{k}\), whatever index the sum begins at. The denominator \(1 - r\) never changes.
A telescoping series is one whose terms can be written as a difference \(a_n = b_n - b_{n + 1}\), so that the partial sums collapse:
Everything in the middle cancels. Convergence is then decided entirely by the single remaining term, which gives an if and only if:
Example: \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\). Partial fractions give \(\frac{1}{n} - \frac{1}{n + 1}\), so \(S_N = 1 - \frac{1}{N + 1} \to 1\).
The difficulty is finding the \(b_n\). Once you have the splitting, the collapse is automatic; partial fractions is the usual route, and where no such splitting exists the technique simply does not apply. That is the whole skill here.
Telescoping and geometric are the only two families of infinite series on this page whose sum you obtain directly. (Section 3's arithmetic formula gives a closed form too, but only for a finite sum — an arithmetic series with \(d \neq 0\) always diverges.) Every convergence test you will meet in the intermediate tutorial answers a strictly weaker question: whether a series converges, never what it converges to.
The two are more closely related than they look. The finite geometric formula of Section 4, \(S_n = a_1 \cdot \frac{1 - r^n}{1 - r}\), is itself proved by telescoping: multiplying \(S_n\) by \((1 - r)\) makes every interior term cancel in exactly this way, leaving \(a_1 - a_1 r^n\).
The harmonic series diverges! Despite the terms going to zero, the partial sums grow without bound (albeit very slowly).
This is the single most important counterexample in series theory, and it is the reason the Divergence Test can never prove convergence.
The claim deserves more than assertion, and the argument is genuinely beautiful — it predates calculus by three centuries. The idea is to group the terms into stretches whose lengths double, and show each stretch contributes at least \(\frac{1}{2}\).
Leave the leading \(1\) aside and group what follows:
Why each stretch clears \(\frac{1}{2}\): the \(k\)-th one runs from \(\frac{1}{2^{k - 1} + 1}\) through \(\frac{1}{2^{k}}\). It therefore holds \(2^{k - 1}\) terms, and every one of them is at least the smallest, \(\frac{1}{2^{k}}\). So the stretch totals at least
Now count. Taking the leading \(1\) together with \(k\) complete stretches accounts for the first \(2^{k}\) terms, so
The partial sums exceed any bound — you need only take \(k\) large enough. Hence the series diverges. The same conclusion follows from the Cauchy Criterion in Section 8, using the stretch from \(n + 1\) to \(2n\); this grouping is that argument made concrete. A fuller write-up is on the Geometric Series and Harmonic Series proof page.
How slowly? Painfully. The partial sums grow like \(S_n \approx \ln n + 0.577\), so reaching a total of 10 takes on the order of \(12{,}000\) terms, and reaching 20 takes hundreds of millions. Divergence says nothing about speed — only that no ceiling exists.
| Series | Converges? | Sum (if convergent) |
|---|---|---|
| \(\sum ar^n\) (\(|r| \lt 1\)) | Yes | \(\frac{a}{1-r}\) |
| \(\sum ar^n\) (\(|r| \geq 1\), \(a \neq 0\)) | No | - |
| \(\sum \frac{1}{n(n+1)}\) (telescoping) | Yes | 1 |
| \(\sum \frac{1}{n}\) (harmonic) | No | - |
Understanding the precise meaning of convergence helps avoid common misconceptions.
Means: \(\lim_{N \to \infty} S_N\) exists and is finite. We may or may not know the exact value.
Means: \(\sum_{n=1}^{\infty} a_n = L\), i.e., we know the specific limit value.
When a series converges to sum \(S\), we can approximate using partial sums:
The remainder \(R_N\) represents how much we're missing by stopping at \(N\) terms. Different convergence tests provide different ways to bound this error.
Everywhere above, the ratio \(r\) has been a fixed number, and \(\sum ar^n\) a single number with it. Nothing forces that. Let the ratio depend on a variable and the same formula becomes considerably more useful: a rule that manufactures a function.
Read it one point at a time. Fix any \(x_0\) in \(D\). Then \(r(x_0)\) is simply a number of modulus less than one, and \(\sum_{n=0}^{\infty} a\,r(x_0)^n\) is an ordinary numerical series of exactly the kind Section 11 handled. It converges, to one specific real number. Nothing about that individual series is new.
Now let \(x\) vary. Every point of \(D\) carries its own numerical series, and therefore its own sum. Collecting those sums — one per point — defines a function \(S : D \to \mathbb{R}\). The series is no longer a number here; it is a construction rule for a function, evaluated pointwise.
The closed form \(\frac{a}{1 - r(x)}\) is then an independent formula for that same function on \(D\) — the same values, obtained without summing anything.
The convergence condition is still \(|r| \lt 1\); it is simply now an inequality in \(x\), and solving it is what produces the domain.
Here \(|r(x)| \lt 1\) reads \(|x| \lt 1\), so \(D = (-1, 1)\) — an interval about the origin.
Here \(|r(x)| \lt 1\) reads \(|x| \gt 1\), so \(D\) is everything outside \([-1, 1]\). A domain need not be an interval at all.
What was described above — a set of results indexed by \(x\) — is the pointwise view, and as a definition it is exactly right. But notice what it deliberately says nothing about: how the family hangs together as \(x\) varies. That silence is where the subtlety lives.
The partial sums themselves are unobjectionable. Each \(S_N(x) = a\sum_{n=0}^{N} r(x)^n\) is a polynomial in \(r(x)\) — continuous, differentiable, entirely harmless. It is tempting to assume the limit function \(S\) simply inherits that good behaviour, and that derivatives may be taken one term at a time:
Swapping a limit with a derivative or an integral is a genuine theorem carrying hypotheses, not bookkeeping. Pointwise convergence on its own is not enough to conclude that \(S\) is even continuous, still less that the interchange above is valid — a family of continuous functions can converge pointwise to a discontinuous limit.
The stronger condition asks that the worst error anywhere on a set \(K\) be driven down all at once. The convergence is uniform on \(K\) if, for every \(\epsilon \gt 0\), some \(N\) makes
One \(N\) serving every \(x\) simultaneously — not an \(N\) chosen separately for each point, which is all the pointwise definition ever promised.
For the geometric case this holds on every compact subset of the domain, which is precisely why term-by-term differentiation and integration of a power series inside its radius of convergence are legitimate. The licence is earned, not free.
The two ideas therefore do different jobs: pointwise convergence defines \(S\), while uniform convergence is what makes \(S\) well behaved. Keeping them apart is the whole point of the distinction. The advanced tutorial develops both definitions in full, along with the Weierstrass M-test that establishes uniform convergence in practice.
Everything in this tutorial rests on a single unifying idea: the limit. A function limit, a sequence limit, and the convergence of a series are not three separate theories — they are three contexts in which the same concept of "approaching a value" is applied.
The sequence limit ($\epsilon$-$N$) studied in this tutorial is tied to the function limit ($\epsilon$-$\delta$) studied in the Comprehensive Guide to Standard Limits by a powerful equivalence — the Heine definition, or sequential characterization of limits.
This means you can test and understand function limits using sequence intuition, even though the two notions are not identical — they are different applications of the same underlying concept. In practice (especially in real analysis), this equivalence is frequently used to prove statements about function limits by working with sequences, where the $\epsilon$-$N$ machinery is often more tractable than a direct $\epsilon$-$\delta$ argument.
| Concept | Domain | Approaching | Rigorous Tool | Key Question |
|---|---|---|---|---|
| Function limit | $\mathbb{R}$ | $x \to a$ (continuous) | $\epsilon$-$\delta$ | Does $f(x)$ stay near $L$ for $x$ near $a$? |
| Sequence limit | $\mathbb{N}$ | $n \to \infty$ (discrete) | $\epsilon$-$N$ | Do terms eventually stay near $L$? |
| Series convergence | $\mathbb{N}$ | Partial sums $S_N$ | $\epsilon$-$N$ on $S_N$ | Does the sequence of partial sums converge? |
Read top to bottom, the table shows the progression of this tutorial: a sequence applies the limit concept discretely, and an infinite series is defined entirely through the limit of its sequence of partial sums (Section 8). The function-limit row connects back to the $\epsilon$-$\delta$ definition developed in the Standard Limits tutorial.