Target Audience: Students who have completed an introduction to series.
Prerequisites: Sequences, partial sums, geometric series, the Divergence Test.
This page lists nine tests, which is more than anyone should memorise. The good news is that they run on only four ideas, and every test below is one of these ideas plus a single inequality.
Terms of mixed sign are handled by testing $\sum |a_n|$ with the first two ideas: if that converges, so does the original (absolute convergence, Section 1). Only when it does not are ideas 3 and 4 needed.
Understanding the distinction between absolute and conditional convergence is crucial because it determines what operations we can safely perform on a series.
A series $\sum a_n$ converges absolutely if the series of absolute values converges:
A series $\sum a_n$ converges conditionally if:
If $\sum |a_n|$ converges, then $\sum a_n$ converges. The converse is not true.
Proof (one line). The trick is to force non-negativity so the Comparison Test of Section 2 becomes available. Since $-|a_n| \leq a_n \leq |a_n|$, we have
So $\sum (a_n + |a_n|)$ converges by comparison with $2\sum |a_n|$. Then $\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|$ is a difference of two convergent series, hence convergent by linearity (Section 11). $\blacksquare$
Converges to $\ln 2$
Diverges (harmonic series)
Therefore, the alternating harmonic series is conditionally convergent.
The comparison tests let us determine convergence by comparing our series to a simpler, well-understood benchmark series.
Let $\sum a_n$ and $\sum b_n$ be series with non-negative terms. If $0 \leq a_n \leq b_n$ for all sufficiently large $n$:
The inequality only has to hold from some index onward — it may fail on any finite stretch at the start. This is the tails principle again, and it matters in practice: a comparison that fails for small $n$ but holds thereafter is still a valid comparison.
Why it works. Suppose the inequality holds from index $M$ on, and $\sum b_n$ converges to $B$. For $N \geq M$ the partial sums of $\sum a_n$ satisfy $S_N \leq (a_1 + \cdots + a_{M - 1}) + B$: a fixed constant plus $B$. They are bounded, and the terms are non-negative, so the Bounded Partial Sums Criterion says $\sum a_n$ converges. The second bullet is the same statement read backwards.
If a "bigger" series converges, the "smaller" one must too. If a "smaller" series diverges, the "bigger" one must also diverge.
Let $\sum a_n$ and $\sum b_n$ be series with positive terms. If:
exists and $0 \lt L \lt \infty$, then both series converge or both diverge.
Special cases: If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
Why it works. Since $a_n / b_n \to L$ with $L \gt 0$, from some index on the ratio lies between $\tfrac{L}{2}$ and $2L$, that is
Each series is therefore squeezed by a constant multiple of the other, and a constant factor never changes whether a series converges. Two applications of Direct Comparison — one in each direction — finish it. The special cases keep just one of the two inequalities.
Good benchmark series include:
Determine convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^4 + n + 1}$.
Strategy: For large $n$, the dominant terms give $\frac{3n^2}{n^4} = \frac{3}{n^2}$.
Compare with $b_n = \frac{1}{n^2}$:
Since $L = 3$ (finite and positive) and $\sum \frac{1}{n^2}$ converges (p-series with $p = 2 \gt 1$), the original series converges.
The Integral Test connects series convergence to the convergence of improper integrals, providing both a convergence criterion and tools for estimating sums.
Let $f(x)$ be a continuous, positive, decreasing function for $x \geq 1$, and let $a_n = f(n)$. Then:
either both converge or both diverge.
The terms $a_n$ are the heights of rectangles of width $1$, and the whole test is one picture drawn twice. Stand each rectangle on $[n, n+1]$ with its height taken at the left end, and because $f$ is decreasing the rectangles poke above the curve; take the height at the right end and they tuck underneath it.
Move a slider to redraw the rectangles.
Left panel: $a_1 + a_2 + \cdots + a_N \geq \int_1^{N+1} f$. Right panel: $a_2 + \cdots + a_N \leq \int_1^{N} f$. The curve is the same in both.
Read together, the two panels trap every partial sum between two integrals:
Why it works. If $\int_1^{\infty} f$ converges, the right-hand inequality bounds every $S_N$ by $a_1 + \int_1^{\infty} f$; the terms are non-negative, so the Bounded Partial Sums Criterion gives convergence. If the integral diverges, the left-hand inequality drives $S_N$ to infinity with it. That is the whole proof — and letting $N \to \infty$ in the two-sided bound gives the familiar
The same picture, started at $x = N$ instead of $x = 1$, is the remainder estimate below.
If $\sum a_n$ converges, the remainder (error when approximating with $N$ terms) satisfies:
This gives us a way to bound how much we're missing!
Using $f(x) = \frac{1}{x^p}$:
Therefore, $\sum \frac{1}{n^p}$ converges if and only if $p \gt 1$.
The p-series is one of the most important reference series. Knowing its convergence behavior is essential for comparison tests.
| Series | Value of p | Behavior |
|---|---|---|
| $\sum \frac{1}{n^2}$ | $p = 2$ | Converges to $\frac{\pi^2}{6}$ |
| $\sum \frac{1}{n^{3/2}}$ | $p = 1.5$ | Converges |
| $\sum \frac{1}{n}$ | $p = 1$ | Diverges (harmonic) |
| $\sum \frac{1}{\sqrt{n}}$ | $p = 0.5$ | Diverges |
With $p = 2 \gt 1$, the series converges.
Partial sums of $\sum_{n=1}^{N} \frac{1}{n^p}$ for $N = 1$ to $50$. The critical threshold is at $p = 1$.
A logarithmic factor in the denominator shrinks the terms, but so gently that it barely shifts the convergence boundary — and a single extra power of $\ln n$ is enough to carry a series across it. These two land on opposite sides:
Diverges
By integral test: $\int_2^{\infty} \frac{1}{x \ln x}\, dx = \ln(\ln x) \big|_2^{\infty} = \infty$
Converges
By integral test: $\int_2^{\infty} \frac{1}{x (\ln x)^2}\, dx = \left[ -\frac{1}{\ln x} \right]_2^{\infty} = \frac{1}{\ln 2}$, finite.
The whole family behaves the same way. The substitution $u = \ln x$ turns $\int_2^{\infty} \frac{dx}{x (\ln x)^p}$ into $\int_{\ln 2}^{\infty} \frac{du}{u^p}$ — a p-series integral in disguise — so $\sum \frac{1}{n (\ln n)^p}$ converges exactly when $p \gt 1$. The logarithm reproduces the p-series boundary one level down.
The Cauchy Condensation Test is an elegant tool particularly suited for testing p-series and logarithmic series.
Let $(a_n)$ be a decreasing sequence of non-negative terms. Then:
We "condense" the series by grouping terms: instead of $a_1, a_2, a_3, \ldots$, we sample at powers of 2: $a_1, a_2, a_4, a_8, \ldots$ and multiply by how many terms each represents.
Why it works — Oresme's grouping, generalised. This is the dyadic grouping that proved the harmonic series diverges, turned into a two-sided estimate. Cut the series into blocks of $2^k$ consecutive terms, from $a_{2^k}$ to $a_{2^{k+1} - 1}$. Because the terms decrease, every term in a block is at most its first term and at least the first term of the next block:
Summing the blocks over $k$ sandwiches the series:
Both sides are series of non-negative terms, so by the Bounded Partial Sums Criterion each converges exactly when its partial sums are bounded — and the sandwich makes the two boundedness conditions equivalent. Nothing about the test is deeper than that.
For $a_n = \frac{1}{n^p}$, apply condensation:
This is a geometric series with ratio $r = 2^{1-p}$.
Note the restriction to $p \gt 0$. Condensation requires a decreasing sequence, and for $p \leq 0$ the terms $\frac{1}{n^p} = n^{-p}$ do not decrease — so the test does not apply at all. Nothing is lost: for $p \leq 0$ the terms fail to approach zero, and the Divergence Test disposes of the series immediately. The two arguments together still give the full result, convergence exactly when $p \gt 1$.
For $\sum \frac{1}{n \ln n}$, let $a_n = \frac{1}{n \ln n}$ and apply condensation:
The condensed series is a multiple of the harmonic series, which diverges. Therefore, $\sum \frac{1}{n \ln n}$ diverges.
The index starts at $k = 1$ here, not $k = 0$ as in the statement of the test. That is forced: the series itself begins at $n = 2$, because $a_1 = \frac{1}{1 \cdot \ln 1}$ divides by zero. Starting the condensation at $k = 1$ samples $a_2, a_4, a_8, \ldots$ and skips the term that does not exist. Shifting where a series starts never affects convergence, only the sum.
The Alternating Series Test (also called the Leibniz Test) applies to series where consecutive terms alternate in sign.
An alternating series has the form:
If $(b_n)$ is a sequence of positive terms such that:
Then $\sum (-1)^{n+1} b_n$ converges.
Why it works. Look at the even and odd partial sums separately. Bracketing the even ones in pairs,
and every bracket is $\geq 0$ because $(b_n)$ decreases, so $(S_{2n})$ is increasing. Bracketing differently,
so it is bounded above too. By the Monotone Convergence Theorem, $S_{2n} \to S$ for some $S$. The odd sums follow for free: $S_{2n+1} = S_{2n} + b_{2n+1} \to S + 0$. Both subsequences converge to $S$, and together they exhaust the sequence, so $S_N \to S$. $\blacksquare$
The same bracketing shows the odd sums decrease to $S$ while the even sums increase to it. So $S$ always sits between two consecutive partial sums, $S_N$ and $S_{N+1}$, which are exactly $b_{N+1}$ apart — and that is the remainder bound below, sign included. The test and its error estimate are one argument, not two.
Condition 1 only has to hold eventually — from some index onward. Finitely many terms contribute a finite amount and cannot change whether a series converges, so a sequence that misbehaves early and then decreases forever still satisfies the test. Like every test on this page, the Alternating Series Test is a statement about the tail of a series.
One of the nicest features of alternating series: we get a free error bound!
If $S = \sum_{n=1}^{\infty} (-1)^{n+1} b_n$ and $S_N$ is the $N$-th partial sum, then:
The error is bounded by the first omitted term. Moreover, the sign of the error is determined by the first omitted term.
Choose $N$ to see the remainder bound.
Partial sums $S_n$ of $\sum (-1)^{n+1} b_n$ alternate around the limit $S$ (dashed), closing in from both sides. The shaded band is $S \pm b_{N+1}$: the remainder bound says $S_N$ (enlarged) must lie inside it, and the picture shows why — $S$ is always trapped between two consecutive partial sums.
We know $\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$
Using $N = 4$ terms: $S_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} \approx 0.5833$
The error bound: $|R_4| \leq b_5 = \frac{1}{5} = 0.2$
Taken as a two-sided bound this gives $0.3833 \leq \ln 2 \leq 0.7833$. But we can do better, because the sign of the error is known: the first omitted term is $+\frac{1}{5}$, so the true value lies above $S_4$, not below it.
Half the width, from information the remainder estimate already gave us. Whenever you use the alternating bound, check the sign of the first omitted term before settling for a symmetric interval.
Both conditions are necessary — and each one fails in its own way. Dropping either is enough to break the test.
Dropping $b_n \to 0$. The series $\sum (-1)^n$ has $b_n = 1 \not\to 0$, so it diverges — caught immediately by the Divergence Test.
Dropping monotonicity. This one is subtler, because the terms still go to zero. Let $b_n = \frac{1}{n}$ for odd $n$ and $b_n = \frac{1}{n^2}$ for even $n$. The signs alternate and $b_n \to 0$, but $(b_n)$ is not decreasing, and
is a convergent piece minus a divergent one, so it diverges to $-\infty$ — the partial sums drift like $-\tfrac{1}{2}\ln N$. Terms tending to zero is not enough on its own.
Both tests handle conditionally convergent series that the Alternating Series Test cannot reach, but they relate to it differently. Dirichlet's Test genuinely generalizes it — the Alternating Series Test is the special case, as shown below. Abel's Test is not a generalization but a different bargain, trading a stronger hypothesis on $\sum a_n$ for a weaker one on $b_n$.
Write $A_n = a_1 + \cdots + a_n$ for the partial sums of one factor. Then for every $N$,
This is integration by parts for sums — $a_n$ plays the derivative, $A_n$ the antiderivative, $b_{n+1} - b_n$ the differential — and it is verified by expanding the right-hand side and cancelling. Its value is that it trades the original series for a boundary term plus a series whose second factor telescopes when $(b_n)$ is monotone.
Both tests below are this identity plus one estimate. It is also what proves Abel's Theorem in the advanced tutorial.
The series $\sum a_n b_n$ converges if:
Why it works. Say $|A_n| \leq M$. The boundary term obeys $|A_N b_N| \leq M b_N \to 0$. In the remaining series, $|A_n (b_{n+1} - b_n)| \leq M (b_n - b_{n+1})$, and $\sum (b_n - b_{n+1})$ telescopes to $b_1$ — so that series converges absolutely by comparison, and the identity hands the convergence back to $\sum a_n b_n$.
If we set $a_n = (-1)^{n+1}$, then the partial sums $A_N$ oscillate between 0 and 1 (bounded). With $b_n$ decreasing to 0, Dirichlet's Test recovers the Alternating Series Test.
The series $\sum a_n b_n$ converges if:
Why it works. Now $A_n \to A$ (so it is bounded) and $b_n \to b$ (monotone and bounded), so the boundary term $A_N b_N \to Ab$. The telescoping sum $\sum |b_{n+1} - b_n|$ equals $|b - b_1|$ because the differences all have one sign, and the same comparison as before finishes the argument.
Show that $\sum_{n=1}^{\infty} \frac{\sin n}{n}$ converges.
Solution: Let $a_n = \sin n$ and $b_n = \frac{1}{n}$.
By Dirichlet's Test, $\sum \frac{\sin n}{n}$ converges.
The Ratio Test and Root Test are powerful tools, especially for series involving factorials, exponentials, or powers.
Let $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$. Then:
That statement assumes the limit exists, which it need not. The general form drops that assumption: $\limsup_{n \to \infty} |a_{n+1}/a_n| \lt 1$ gives absolute convergence, and $\liminf_{n \to \infty} |a_{n+1}/a_n| \gt 1$ gives divergence.
Note also where the test is useless: for rational functions of $n$, such as a p-series, the ratio tends to 1 regardless of convergence. Reach for comparison instead.
A sequence need not have a limit, but it always has a largest and a smallest value that it keeps returning to. For a bounded sequence $(x_n)$, set
The inner quantity $\sup_{k \geq n} x_k$ is the supremum of the tail from $n$ onward. Discarding a term can only lower a supremum, so these tail suprema form a decreasing sequence, bounded below — and the Monotone Convergence Theorem of the basic tutorial guarantees that its limit exists. The same argument, upside down, handles the infimum. That is why $\limsup$ and $\liminf$ always exist where a plain limit may not. (For an unbounded sequence the value is taken to be $\pm\infty$.)
Example: $x_n = (-1)^n + \frac{1}{n}$ has no limit. Its even terms approach $1$ from above and its odd terms approach $-1$ from above, so $\limsup x_n = 1$ and $\liminf x_n = -1$.
Always $\liminf x_n \leq \limsup x_n$, with equality exactly when $\lim x_n$ exists — and then all three agree. In the Ratio Test above, $\limsup \lt 1$ says the ratios are eventually all below some $r \lt 1$, which is what a comparison with a geometric series needs; $\liminf \gt 1$ says they are eventually all above $1$, so the terms cannot shrink to zero.
Let $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$. Then:
The general form uses $L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$, which always exists. That is the version in which the Root Test is strictly stronger than the Ratio Test.
Why they work. Both tests are a comparison with a geometric series. If $L \lt 1$, choose any $r$ with $L \lt r \lt 1$. For the Ratio Test, from some index $N$ on the ratios are below $r$, so each term is at most $r$ times the one before:
The tail is dominated by the convergent geometric series $|a_N| \sum r^k$, so $\sum |a_n|$ converges by Direct Comparison. For the Root Test the same $r$ gives $|a_n| \leq r^n$ directly, from some index on.
If $L \gt 1$ the inequalities reverse: for the Ratio Test the terms eventually grow in absolute value, and for the Root Test $|a_n| \geq 1$ infinitely often. Either way $a_n \not\to 0$, and the Divergence Test finishes. At $L = 1$ neither comparison is available — which is exactly why the tests say nothing there.
They are not equals, and it is worth knowing which way the relationship runs.
The Root Test is strictly stronger than the Ratio Test.
If the Root Test dominates, why keep the Ratio Test? Because it is usually far easier to compute. With factorials the ratio $\frac{a_{n+1}}{a_n}$ collapses through cancellation, while an $n$-th root of a factorial is unpleasant. Strength and convenience point in opposite directions, so both stay in the toolkit.
The gap between them is not hypothetical. When a series has coefficients that vanish infinitely often, the ratio is undefined for infinitely many $n$ and the Ratio Test never applies, while the Root Test proceeds without difficulty — worked through on the advanced page using $\sum x^{2n}$.
| Series Contains | Preferred Test |
|---|---|
| Factorials ($n!$) | Ratio Test |
| Powers with $n$ in exponent ($a^n$, $n^n$) | Root Test |
| Products of factorials and powers | Ratio Test |
| $n$-th power of expression with $n$ | Root Test |
Test convergence of $\sum_{n=1}^{\infty} \frac{n!}{n^n}$.
Since $L = \frac{1}{e} \lt 1$, the series converges.
Test convergence of $\sum_{n=1}^{\infty} \left(\frac{2n+1}{3n+2}\right)^n$.
Since $L = \frac{2}{3} \lt 1$, the series converges.
When $L = 1$, the Ratio and Root Tests tell us nothing. The p-series family makes this vivid: every p-series gives $L = 1$, yet the family contains both convergent and divergent members.
Same $L$, opposite verdicts, so $L = 1$ cannot possibly carry the answer. Use a different test.
Raabe's Test is a refinement of the Ratio Test, useful when the Ratio Test is inconclusive (i.e., when $L = 1$).
Let $R = \lim_{n \to \infty} n\left(1 - \frac{|a_{n+1}|}{|a_n|}\right) = \lim_{n \to \infty} n\left(\frac{|a_n| - |a_{n+1}|}{|a_n|}\right)$. Then:
What $R$ measures. For the p-series itself, $a_{n+1}/a_n = \left(\frac{n}{n+1}\right)^p$ and $n\left(1 - \left(\frac{n}{n+1}\right)^p\right) \to p$. So Raabe's Test reads the ratio one order more finely than the Ratio Test — as $1 - \frac{R}{n} + \cdots$ — and compares the series with $\sum \frac{1}{n^R}$, which is why the verdict flips at $R = 1$ just as the p-series does at $p = 1$.
Raabe's Test is particularly useful for series involving gamma functions, beta functions, or ratios of products (e.g., $\frac{(2n)!}{4^n (n!)^2}$).
Test convergence of $\sum_{n=1}^{\infty} \frac{(2n)!}{4^n (n!)^2}$.
Step 1: Compute the ratio:
As $n \to \infty$: $\frac{a_{n+1}}{a_n} \to 1$. The Ratio Test is inconclusive.
Step 2: Apply Raabe's Test:
Since $R = \frac{1}{2} \lt 1$, the series diverges.
With so many tests available, choosing the right one is an art. Here's a systematic approach.
| If You See... | Try... |
|---|---|
| $(-1)^n$ or alternating signs | Alternating Series Test |
| $n!$, $(2n)!$, factorials | Ratio Test |
| $n^n$, $a^n$, $\left(\frac{\cdot}{\cdot}\right)^n$ | Root Test |
| $\frac{1}{n^p}$, polynomials in $n$ | p-Series / Comparison Tests |
| $\ln n$ in denominator | Integral Test or Cauchy Condensation |
| Ratio Test gives $L = 1$ | Raabe's Test |
Understanding what operations preserve convergence is essential for working with series.
If $\sum a_n$ and $\sum b_n$ both converge:
Linearity does not extend to products: $\sum a_n b_n$ is not $\left(\sum a_n\right)\left(\sum b_n\right)$. Take $a_n = b_n = \frac{1}{2^n}$ from $n = 1$. Each series sums to $1$, so the product of the sums is $1$ — but $\sum a_n b_n = \sum \frac{1}{4^n} = \frac{1}{3}$.
Multiplying two sums means multiplying every term by every term, the way polynomials are multiplied, and collecting the results by total index. That is the Cauchy product, and even it needs a hypothesis before its sum is the product of the sums.
Grouping consecutive terms (without reordering) inserts brackets without moving anything:
If the original series converges, every such regrouping converges to the same sum. The reason is simple: the partial sums of the regrouped series are a subsequence of the original partial sums, and a subsequence of a convergent sequence has the same limit.
That is the whole guarantee, and it runs in one direction only.
A convergent regrouping tells you nothing about the original series. Bracketing can manufacture convergence that was never there:
Same terms, same order, brackets added — and the verdict changes. Bracketing hides the oscillation by only ever letting you look at the even-numbered partial sums. This is exactly the move behind the bogus "proof" that $1 - 1 + 1 - 1 + \cdots = 0$. Group a series you have already shown converges; never group one in order to show it converges.
If $\sum a_n$ is conditionally convergent, then for any real number $S$ (or $\pm\infty$), there exists a rearrangement of the series that converges to $S$.
In other words: conditionally convergent series can be rearranged to sum to anything!
The alternating harmonic series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ sums to $\ln 2 \approx 0.693$.
A rearrangement summing to $\frac{3}{2}\ln 2$:
Take two positive terms, then one negative term:
This rearrangement converges to $\frac{3}{2}\ln 2 \approx 1.039$!
Key insight: This is why absolute convergence matters—absolutely convergent series can be rearranged freely without changing the sum.
Never rearrange terms of a conditionally convergent series unless you know exactly what you're doing. The sum can change arbitrarily!
In practice, we often need to know: "How many terms do I need for a certain accuracy?"
Problem: Approximate $\sum_{n=1}^{\infty} \frac{1}{n^3}$ to within $10^{-6}$.
Solution: Using the integral test remainder:
We need $\frac{1}{2N^2} \lt 10^{-6}$:
So we need approximately 708 terms to guarantee 6 decimal place accuracy.
The two-sided remainder bound does more than certify accuracy: it says where the tail is. Adding just $\int_N^{\infty} f$ to $S_N$ would land on the upper end of the bracket, and always overshoot. Aim for the middle instead:
The guaranteed error is half the width of the bracket, and in practice the estimate does far better than the guarantee, because the tail behaves very much like the integral. For $\sum \frac{1}{n^3}$ with $N = 10$: $S_{10}$ alone is off by $4.5 \times 10^{-3}$; $S_{10} + \int_{10}^{\infty}$ is off by $4.8 \times 10^{-4}$, on the high side; the midpoint is off by $4 \times 10^{-5}$, with a guarantee of $4.4 \times 10^{-4}$.
Applied to the $708$-term computation above, the midpoint estimate is within $2 \times 10^{-12}$ of the true sum — eleven correct decimal places from the same $708$ terms that the bare partial sum needed just to reach six.
If we can bound $a_n$ by a geometric series $|a_n| \leq C r^n$ for $n \geq N$, then:
This can give very tight bounds for series that decay geometrically.
The Ratio Test's proof (Section 8) is a geometric comparison, so it carries an error bound with it. If $|a_{n+1}| \leq r\, |a_n|$ for every $n \geq N + 1$, with $r \lt 1$, then the tail is dominated by $|a_{N+1}| (1 + r + r^2 + \cdots)$:
For $\sum \frac{1}{n!}$ the ratios are $\frac{1}{n+1}$, so $r = \frac{1}{N+2}$ works from index $N + 1$ on. With $N = 10$ the bound is $2.733 \times 10^{-8}$ against a true remainder of $2.731 \times 10^{-8}$ — tight to three figures, because the ratios shrink so fast that the first omitted term is almost the whole tail.