Target Audience: Students with solid foundations in series convergence tests.
Prerequisites: Intermediate series theory, convergence tests, absolute/conditional convergence.
Sections 1–8 are a connected development: what a power series is, why it has a radius, what licenses differentiating it term by term, and the two notions — uniform convergence and absolute convergence of double series — that make the licences honest. Read them in order; each one uses the last.
Sections 9–15 change gear. They are a tour of where series go next — Fourier series, generating functions, infinite products, complex power series, summability, asymptotics and the special functions — each of which is a course in its own right. Those sections state results and show what they are for; they do not prove them, and they can be read in any order.
A power series is the backbone of advanced analysis, connecting infinite series to functions in a profound way. The basic tutorial ended by letting the ratio of a geometric series depend on $x$ and reading the result as a function, defined one point at a time. Power series are the canonical case of that construction — the terms are monomials in $x$ — and this page is about what more can be said when they are.
A power series centered at $a$ is an infinite series of the form:
where $c_n$ are the coefficients and $a$ is the center.
Every power series has a radius of convergence $R$ such that the series:
Nothing in the definition says the set of $x$ where a power series converges should be an interval about the centre — yet it always is, and the reason is one line of geometric comparison, usually called Abel's lemma. Take the centre at $a = 0$ and suppose the series converges at some $x_0 \neq 0$. Then its terms tend to zero, so they are bounded: $|c_n x_0^n| \leq M$ for all $n$. For any $x$ closer to the centre,
The right-hand side is a convergent geometric series, so $\sum c_n x^n$ converges absolutely. Convergence at one point therefore forces absolute convergence at every point nearer the centre; divergence at one point forces divergence at every point farther out. The convergence set is an interval, and $R$ is simply the supremum of $|x - a|$ over the points where the series converges — with $R = 0$ and $R = \infty$ allowed. That is the whole content of the three bullets above.
Using the Ratio Test, provided the coefficients are eventually nonzero and the limit exists (possibly as $\infty$):
Using the Root Test (Cauchy-Hadamard formula):
with the conventions $1/0 = \infty$ and $1/\infty = 0$. Note the $\limsup$ (defined in the intermediate tutorial): unlike an ordinary limit it always exists, so Cauchy-Hadamard always applies. The ratio formula above does not.
The two extremes are real. $\sum n!\, x^n$ has $\sqrt[n]{n!} \to \infty$, so $R = 0$: it converges at the centre and nowhere else. $\sum x^n / n!$ has $\sqrt[n]{1/n!} \to 0$, so $R = \infty$: it converges for every $x$. Both are read off the same formula.
The two formulas are not interchangeable, and the difference appears as soon as coefficients vanish infinitely often. Take
Here $c_n = 1$ for even $n$ and $c_n = 0$ for odd $n$. The ratio $\left|\frac{c_{n+1}}{c_n}\right|$ is therefore undefined for every odd $n$ — the denominator is zero — so the ratio formula never gets started.
Cauchy-Hadamard has no such difficulty. The values of $\sqrt[n]{|c_n|}$ are $1$ (even $n$) and $0$ (odd $n$), so the $\limsup$ is $1$, giving $R = 1$.
Confirmed independently: $\sum x^{2n} = \sum (x^2)^n$ is geometric in $x^2$, so it converges exactly when $|x^2| \lt 1$, that is $|x| \lt 1$ — radius $1$, as Cauchy-Hadamard said. This is precisely why the Root Test is strictly stronger than the Ratio Test.
The interval of convergence is the set of all $x$ for which the series converges. It always includes $(a-R, a+R)$, but may include one, both, or neither endpoint.
Radius of Convergence: R = 1
Interval: $(-1, 1)$
Partial sums of the power series for various values of $x$.
Find the radius and interval of convergence for $\sum_{n=1}^{\infty} \frac{x^n}{n}$.
Step 1: Apply the Ratio Test:
So $R = 1$ and the series converges absolutely for $|x| \lt 1$.
Step 2: Check endpoints:
Interval of convergence: $[-1, 1)$
Within their radius of convergence, power series define smooth functions that can be differentiated and integrated term-by-term.
If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ has radius of convergence $R \gt 0$, then:
The derivative has the same radius of convergence $R$.
If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ has radius of convergence $R \gt 0$, then:
The integral also has radius of convergence $R$ (though endpoint behavior may differ).
Differentiating an infinite sum term by term is an interchange of two limits, and the basic tutorial was careful to say that pointwise convergence does not license it. Here is what does. The derivative series has the same radius: its coefficients are $n c_n$, and $\sqrt[n]{n} \to 1$, so Cauchy-Hadamard returns the same $R$. Inside that radius, on any closed interval $[a - r, a + r]$ with $r \lt R$, the derivative series converges uniformly — the Weierstrass M-Test of Section 7 shows this — and the Differentiation Interchange theorem of Section 6 then applies on that interval. Every point of the open interval $|x - a| \lt R$ lies inside some such $[a - r, a + r]$, which is why the conclusion holds throughout the open interval and never at an endpoint.
Integration is the same story with the Integration Interchange theorem. This is the licence the basic tutorial said had to be earned; Sections 6 and 7 are where it is earned.
Starting with the geometric series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| \lt 1$:
Differentiate:
Integrate:
We want $\sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^n$.
From differentiating the geometric series:
Multiply by $x$:
Substitute $x = \frac{1}{2}$:
Abel's Theorem addresses what happens at the boundary of convergence—a delicate situation where the series might converge but standard analysis techniques don't apply directly.
Let $f(x) = \sum_{n=0}^{\infty} c_n x^n$ have radius of convergence $R \gt 0$. If the series converges at $x = R$ (i.e., $\sum c_n R^n$ converges), then:
In other words, the power series function is continuous from the left at the boundary.
Abel's Theorem allows us to evaluate limits by substituting boundary values into convergent series, even when the function itself might not be defined there in a traditional sense.
Integrating the geometric series for $\frac{1}{1+x}$ term by term, as in Section 2, gives $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$ for $|x| \lt 1$ — and no further, since term-by-term integration only reaches the open interval.
The series converges at $x = 1$ (alternating harmonic series).
By Abel's Theorem:
More generally, if $\sum a_n$ converges to $S$, then:
This connects power series limits to ordinary series sums.
The limit $\lim_{x \to 1^-} \sum a_n x^n$ may exist even when $\sum a_n$ diverges! This is the basis for Abel summation (covered in Section 13).
A concrete case makes the trap clear. Take $a_n = (-1)^n$:
The left-hand limit exists and equals $\frac{1}{2}$ — yet $\sum (-1)^n = 1 - 1 + 1 - \cdots$ does not converge at all. So a limit existing does not make the endpoint series convergent. Abel's Theorem runs in one direction only: it needs convergence at the endpoint as a hypothesis, and cannot be used to establish it.
Taylor series represent functions as power series using derivatives at a single point.
For $f$ infinitely differentiable at $a$, the Taylor series of $f$ centered at $a$ is the series
A Maclaurin series is a Taylor series centered at $a = 0$.
Note carefully what is not being claimed: that this series equals $f(x)$. It is built from $f$, but whether it converges back to $f$ is a separate question. Writing $f(x) = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$ is a theorem to be earned at each $x$, not part of the definition — and it holds precisely where the remainder satisfies $R_n(x) \to 0$. See When Taylor Series Fail below for a function where it fails everywhere except a single point.
| Function | Maclaurin Series | Convergence |
|---|---|---|
| $e^x$ | $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ | All $x$ ($R = \infty$) |
| $\sin x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ | All $x$ ($R = \infty$) |
| $\cos x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ | All $x$ ($R = \infty$) |
| $\ln(1+x)$ | $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$ | $(-1, 1]$ |
| $(1+x)^\alpha$ | $\sum_{n=0}^{\infty} \binom{\alpha}{n} x^n$ | $|x| \lt 1$ for non-integer $\alpha$; all $x$ if $\alpha$ is a non-negative integer, where the series terminates |
The table asserts that $e^x$, $\sin x$ and $\cos x$ equal their Maclaurin series everywhere. By the definition's own warning, that is a theorem — and it follows from the remainder bound below. For $\sin$ and $\cos$ every derivative is bounded by $K = 1$; for $e^x$ every derivative is $e^x$ itself, bounded by $K = e^{|x|}$ on the interval from $0$ to $x$. Either way,
because factorials outgrow every power: at $x = 10$ the bound is already $10^{61}/61! \approx 2 \times 10^{-23}$ at $n = 60$. The remainder vanishes at every $x$, so the series converges to the function at every $x$. Compare $\ln(1+x)$, whose derivatives $\pm (n-1)!/(1+x)^n$ grow factorially and defeat the same argument outside $|x| \lt 1$.
That half-open interval is doing real work, and no single argument establishes all of it. Three separate results are involved, one per piece:
The same three-part pattern recovers $\arctan 1 = \frac{\pi}{4}$ from the series for $\arctan x$. Whenever a table lists a half-open interval, this is the machinery underneath it.
If $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$:
where $T_n(x)$ is the $n$-th degree Taylor polynomial and:
for some $\xi$ between $a$ and $x$.
The existence of that $\xi$ is the whole content of the theorem, and it follows from the Mean Value Theorem applied to a carefully chosen auxiliary function — worked through in full on the Taylor's Theorem with Lagrange Remainder proof page.
The theorem is not merely descriptive — it is the tool that lets you decide how many terms you need, before computing any of them. The obstacle is that $\xi$ is unknown; the resolution is that you never need it, only a bound on the derivative over the whole interval.
If $|f^{(n+1)}| \leq K$ on the interval between $a$ and $x$, then
In practice this is three steps:
For $\sin x$ every derivative is $\pm\sin$ or $\pm\cos$, so $K = 1$ always — which is what makes it such a convenient test case. Approximate $\sin(0.5)$ by $x - \frac{x^3}{6}$, the degree-3 polynomial:
The actual error is $0.00026$ — about ten times smaller than the bound. That looks like the estimate being wasteful, but it is not luck, and it is worth understanding why.
$\sin$ has no $x^4$ term, so its degree-3 and degree-4 polynomials are identical. The same approximation therefore also carries the $n = 4$ bound:
which matches the true error almost exactly. The lesson: do not expect an order of magnitude of slack in general. Expect the bound to be tight for the first $n$ whose term actually appears in the series — and loose whenever you have accidentally quoted it at an $n$ whose term is missing.
Taylor polynomial (blue) vs. actual function (green). More terms = better approximation near center.
Not every smooth function equals its Taylor series!
The classic example: $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$.
All derivatives at $0$ are $0$, so the Maclaurin series is $0$ everywhere, but the function is not identically zero. This function is smooth but not analytic.
How do we multiply two infinite series? The Cauchy product provides the answer.
Given $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$, their Cauchy product is:
This is analogous to multiplying polynomials: $c_n$ collects all products $a_k b_j$ with $k + j = n$.
If $\sum a_n$ converges absolutely to $A$ and $\sum b_n$ converges (not necessarily absolutely) to $B$, then their Cauchy product converges to $AB$.
If both $\sum a_n$ and $\sum b_n$ converge absolutely, then their Cauchy product also converges absolutely to $\left(\sum a_n\right)\left(\sum b_n\right)$.
Let $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$. The Cauchy product of $e^x$ with itself:
So indeed $e^x \cdot e^x = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = e^{2x}$. ✓
If both series are only conditionally convergent, the Cauchy product may diverge!
Example: The Cauchy product of $\sum \frac{(-1)^n}{\sqrt{n+1}}$ with itself diverges, even though the original series converges (conditionally).
Uniform convergence is the key concept that allows us to swap limits with integrals, derivatives, and infinite sums.
A sequence of functions $(f_n)$ converges pointwise to $f$ on a set $S$ if:
The rate of convergence may vary with $x$.
$(f_n)$ converges uniformly to $f$ on $S$ if:
The key difference: $N$ works for all $x$ simultaneously.
Equivalently, the worst error over the set goes to zero: $\sup_{x \in S} |f_n(x) - f(x)| \to 0$. That is the form the basic tutorial used, and the two say the same thing.
If each $f_n$ is continuous and $f_n \to f$ uniformly, then $f$ is continuous.
If $f_n \to f$ uniformly on $[a,b]$, then: $\int_a^b f_n \to \int_a^b f$
If $(f_n)$ converges pointwise to $f$, each $f_n'$ exists and is continuous, and $(f_n')$ converges uniformly to some function $g$, then:
For series $\sum f_n(x)$, let $S_N(x) = \sum_{n=1}^{N} f_n(x)$. The series converges uniformly if the partial sums $S_N$ converge uniformly.
$f_n(x) = x^n$ on $[0, 1]$ converges pointwise to:
This is not uniform convergence: each $f_n$ is continuous, but the limit $f$ is discontinuous!
The Weierstrass M-Test is the primary tool for proving uniform convergence of series of functions.
Let $(f_n)$ be a sequence of functions on a set $S$. If there exist constants $M_n$ such that:
Then $\sum f_n(x)$ converges uniformly and absolutely on $S$.
The $M_n$ are "majorants" that bound each term uniformly across all $x$. If the sum of these bounds converges, so does the original series—uniformly.
Show that $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges uniformly on $[-1, 1]$.
Solution: For $x \in [-1, 1]$:
Since $\sum \frac{1}{n^2}$ converges (p-series with $p = 2 \gt 1$):
By the Weierstrass M-Test, the series converges uniformly on $[-1, 1]$. ✓
A power series $\sum c_n (x-a)^n$ with radius $R$ converges uniformly on any closed interval $[a-r, a+r]$ where $0 \lt r \lt R$.
Why? For $|x - a| \leq r$:
Since $r \lt R$, the series $\sum |c_n| r^n$ converges (inside the radius of convergence), so the M-Test applies.
The M-Test proves uniform and absolute convergence. If a series converges uniformly but not absolutely (conditionally), the M-Test cannot be used directly. For such cases, consider Dirichlet's test for uniform convergence.
A double series involves summing over two indices. The key question: when can we change the order of summation?
A double series has terms $a_{m,n}$ indexed by two integers — an infinite array. There are two quite different things one might mean by its sum, and the section's title names both.
The double sum $\sum_{m,n} a_{m,n}$ treats the array as one collection of numbers with no order. For non-negative terms it is defined as the supremum of the sums over finite subsets of the array; since every finite subset lies inside some rectangle, any way of exhausting the plane — growing squares, rows, columns, diagonals — produces the same value. For terms of mixed sign the series is said to converge absolutely when $\sum_{m,n} |a_{m,n}| \lt \infty$ in this sense, and its double sum is then $\sum a_{m,n}^{+} - \sum a_{m,n}^{-}$.
An iterated series such as $\sum_{m=1}^{\infty} \left( \sum_{n=1}^{\infty} a_{m,n} \right)$ is a different object: an inner sum computed first, then an outer sum of the results, with the order fixed by the notation. The whole question of this section is when the iterated sums agree with each other and with the double sum.
Tonelli. If every $a_{m,n} \geq 0$, the double sum and both iterated sums are equal — all three finite, or all three $+\infty$.
Fubini. If $\sum_{m,n} |a_{m,n}|$ converges (absolute convergence of the double series), then every row and every column converges, both iterated series converge, and
Tonelli is applied to $|a_{m,n}|$ to check the hypothesis of Fubini — which is exactly how the example below proceeds.
Compute $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2 n^2}$.
Solution: By Fubini (the double series converges absolutely):
Without absolute convergence, the two iterated sums can genuinely disagree — and the following example pins it down exactly, with no hedging required.
Let $a_{m,n} = 1$ when $n = m$, let $a_{m,n} = -1$ when $n = m+1$, and let every other entry be $0$. So the array carries $+1$ down the diagonal and $-1$ down the superdiagonal.
The single unmatched $+1$ in the first column is the entire discrepancy. Summing rows first pairs it off against a $-1$; summing columns first strands it. This is Riemann rearrangement one dimension up — without absolute convergence, the order of summation carries the answer.
Everything up to this point has been developed with proofs or pointers to them. The remaining sections are a survey: each states the central results of a subject that grows out of series, shows what they are for, and stops. Treat them as a map of where to go next rather than as the territory.
Fourier series expand periodic functions in terms of sines and cosines—the building blocks of periodic phenomena.
For a function $f$ with period $2\pi$, the Fourier series is:
where the Fourier coefficients are:
The key insight: $\{\cos(nx), \sin(nx)\}$ form an orthogonal basis for periodic functions — in the $L^2$ sense, meaning for square-integrable functions, with the partial sums converging in mean square. The Fourier coefficients are projections onto this basis.
Similar relations hold for $\sin$ and mixed $\sin$-$\cos$ products.
The smoothness hypothesis is not decorative. Continuity alone does not guarantee that a Fourier series converges pointwise: du Bois-Reymond showed in 1873 that a continuous $2\pi$-periodic function can have a Fourier series that diverges at a point. Some regularity beyond continuity — piecewise smoothness as above, or bounded variation — is what buys convergence, and statements that begin "at points of continuity" are silently assuming it.
The periodicity condition in the uniform bullet is easy to overlook and cannot be dropped either. Continuity on $[-\pi, \pi]$ is not enough: if the endpoint values disagree, the periodic extension has a jump at $\pm\pi$, and the convergence there is not uniform no matter how smooth $f$ looks on the open interval. It is the extension, not the function on one period, that the Fourier series is approximating.
Near discontinuities, Fourier partial sums exhibit overshoot of about 9%.
This overshoot doesn't disappear as more terms are added—it just becomes narrower and moves closer to the discontinuity.
The exact constant comes from the sine integral. The partial sums peak at
times half the jump. So the overshoot past the limiting value is about 17.9% of the half-jump — which is the same thing as 8.95% of the full jump, the figure usually quoted as "about 9%".
Watch the baseline. Both percentages appear in the literature and they differ by a factor of two; a statement of the Gibbs constant is meaningless without saying which jump it is measured against. For the square wave below, running between $-1$ and $1$, the partial sums peak at roughly $1.1790$: an excess of $0.1790$ over the limit $1$, which is $17.9\%$ of that limit and $8.95\%$ of the total jump of $2$.
The square wave $f(x) = \begin{cases} 1 & 0 \lt x \lt \pi \\ -1 & -\pi \lt x \lt 0 \end{cases}$ has Fourier series:
The $\sim$ is deliberate, and this example shows why the definition above used it too. At $x = 0$ and $x = \pm\pi$ every term $\sin(nx)$ vanishes, so the series sums to $0$ — not to $\pm 1$. That is exactly the "average of the left and right limits" behaviour listed above, since $\tfrac{1}{2}(1 + (-1)) = 0$. The series represents $f$ everywhere except at its jumps, where it splits the difference instead.
Generating functions encode sequences as power series, transforming discrete problems into continuous analysis.
The ordinary generating function (OGF) for a sequence $(a_n)$ is:
Generating functions are usually handled as formal power series: the identities in the table below are statements about coefficients, obtained by algebra on the symbols, and they hold whether or not the series converges anywhere. Wherever it does converge, the same identities hold for the functions too — and only then may values be substituted for $x$.
| Sequence Operation | GF Operation |
|---|---|
| $(a_0, a_1, a_2, \ldots)$ | $A(x)$ |
| $(0, a_0, a_1, \ldots)$ (shift right) | $xA(x)$ |
| $(a_1, a_2, a_3, \ldots)$ (shift left) | $\frac{A(x) - a_0}{x}$ |
| $(a_n + b_n)$ | $A(x) + B(x)$ |
| $\sum_{k=0}^{n} a_k b_{n-k}$ (convolution) | $A(x) \cdot B(x)$ |
The Fibonacci sequence satisfies $F_n = F_{n-1} + F_{n-2}$ with $F_0 = 0$, $F_1 = 1$.
Let $F(x) = \sum_{n=0}^{\infty} F_n x^n$. From the recurrence:
Solving:
Using partial fractions and expanding:
where $\phi = \frac{1+\sqrt{5}}{2}$ (golden ratio) and $\psi = \frac{1-\sqrt{5}}{2}$.
As a function, $F(x)$ converges for $|x| \lt 1/\phi \approx 0.618$ — the distance to the nearer zero of $1 - x - x^2$, which is where the closed form has its pole (Section 12 explains why that is always the radius). The sequence itself, and the golden-ratio limit of $F_{n+1}/F_n$ that Binet's formula makes obvious, were met in the basic tutorial.
Infinite products generalize infinite sums and have deep connections to series through logarithms.
An infinite product is:
We say the product converges if this limit exists and is nonzero.
For positive terms $a_n \gt 0$, the product $\prod a_n$ converges if and only if $\sum \ln a_n$ converges.
This transforms products into sums!
If $a_n \gt -1$ for all $n$, then $\prod (1 + a_n)$ converges absolutely if and only if $\sum a_n$ converges absolutely.
(For small $a_n$: $\ln(1 + a_n) \approx a_n$)
One of the most beautiful formulas in mathematics connects series and products over primes:
The domain restriction is not decorative: both sides diverge for $\operatorname{Re}(s) \leq 1$, and at $s = 1$ the series is the harmonic series. (The zeta function is defined beyond this strip by analytic continuation, but the product formula itself holds only where both sides converge.)
This product representation is the gateway to the deep connections between analysis and number theory.
Extending power series to complex numbers reveals why real series behave the way they do.
Consider $f(x) = \frac{1}{1+x^2}$. Its Maclaurin series:
has radius of convergence $R = 1$. But wait—$f(x)$ is perfectly smooth for all real $x$!
The explanation: In the complex plane, $f(z) = \frac{1}{1+z^2}$ has poles at $z = \pm i$. These complex singularities are at distance 1 from the origin.
The radius of convergence equals the distance to the nearest singularity in the complex plane.
A function $f$ is analytic at a point $z_0$ if it equals its Taylor series in some neighborhood of $z_0$.
Complex-differentiable functions are automatically analytic (a remarkable theorem!).
Analytic continuation extends a function beyond its original domain while preserving analyticity.
Example: The geometric series $\sum z^n$ converges only for $|z| \lt 1$, but it equals $\frac{1}{1-z}$, which is defined everywhere except $z = 1$.
The function $\frac{1}{1-z}$ is the analytic continuation of $\sum z^n$.
This follows from comparing Taylor series:
The middle step uses $i^n$ cycling through $1, i, -1, -i$ and then sorts the even and odd terms into two separate series. That sorting is a rearrangement, and it is legitimate only because $\sum |(ix)^n / n!| = \sum |x|^n / n! = e^{|x|}$ converges — absolute convergence is what permits rearranging without changing the sum.
Summability methods assign values to divergent series in a consistent, useful way.
Some divergent series arise naturally in physics and other applications. Rather than dismissing them, summability theory asks: "Is there a sensible value to assign?"
The Cesàro sum of $\sum a_n$ is the limit of the averages of partial sums:
where $S_n = \sum_{k=0}^{n} a_k$.
Consider $1 - 1 + 1 - 1 + \cdots$ (Grandi's series). The partial sums oscillate: $1, 0, 1, 0, \ldots$
The averages:
These converge to $\frac{1}{2}$. So $(C,1)\text{-}\sum (-1)^n = \frac{1}{2}$.
The Abel sum is:
when this limit exists.
Summability assigns values consistently but differently from ordinary convergence. The statement "$1 - 1 + 1 - 1 + \cdots = \frac{1}{2}$" requires specifying the summation method.
Asymptotic analysis describes how functions and sequences behave for large arguments, providing tools for estimating convergence and growth.
We write $f(n) = O(g(n))$ as $n \to \infty$ if there exist constants $C, N_0$ such that:
"$f$ grows no faster than $g$."
We write $f(n) = o(g(n))$ as $n \to \infty$ if:
"$f$ grows strictly slower than $g$."
We write $f(n) \sim g(n)$ if:
"$f$ and $g$ are asymptotically equal."
| Relation | Meaning |
|---|---|
| $n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$ | Stirling's approximation |
| $\sum_{k=1}^{n} \frac{1}{k} \sim \ln n$ | Harmonic series growth |
| $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$ as $x \to 0$ | Taylor expansion with error — note the regime is $x \to 0$, not $n \to \infty$; the same notation, read at the other end |
Example: Does $\sum \frac{1}{n^2 + n}$ converge?
Solution: For large $n$: $\frac{1}{n^2 + n} \sim \frac{1}{n^2}$.
Since $\sum \frac{1}{n^2}$ converges (p-series) and the terms are asymptotically equivalent, the original series converges by the Limit Comparison Test.
Many important functions in mathematics and physics are defined by their series representations.
Converges for all $x$. The exponential is the unique function satisfying $f'(x) = f(x)$ with $f(0) = 1$.
Converges for $-1 \lt x \leq 1$. Note the asymmetric interval!
where the generalized binomial coefficient is:
For non-integer $\alpha$, converges for $|x| \lt 1$.
Converges for $\text{Re}(s) \gt 1$. Special values include:
Bessel functions of the first kind arise in cylindrical problems:
These solve Bessel's differential equation and appear throughout physics.
While usually defined by an integral, the Gamma function connects to series via:
where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant.