Math Core

Power series can be treated almost exactly like polynomials: you can differentiate them term by term, integrate them term by term, and substitute into them. These moves let you build new series from old ones without computing a single high-order derivative, and they let you approximate integrals that have no elementary antiderivative.

Differentiating and integrating term by term

Calculus with power series

If $f(x) = \displaystyle\sum_{n=0}^{\infty} c_n(x - a)^n$ has radius of convergence $R > 0$ , then on $|x - a| < R$ :

f'(x) = \sum_{n=1}^{\infty} n\,c_n (x - a)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x - a)^{n+1}.

Both new series have the same radius of convergence $R$ . The behavior at the endpoints can change, so endpoints must be rechecked.

As a quick check, differentiate the series for $\sin x$ term by term:

\frac{d}{dx}\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right) = 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \cdots = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x.

Deriving new series from the geometric series

The geometric series $\dfrac{1}{1 - x} = \sum x^n$ is the starting point for a whole family of series.

Substitution. Replace $x$ by $-x$ : $\dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots$ . Replace $x$ by $-x^2$ : $\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots$ . Both hold for $|x| < 1$ .

Integration. Integrate $\dfrac{1}{1+x}$ from $0$ to $x$ :

\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}.

Integrate $\dfrac{1}{1 + x^2}$ from $0$ to $x$ :

\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}.

The radius stays $1$ , but now the endpoints behave differently: the $\ln(1+x)$ series converges at $x = 1$ (giving $\ln 2 = 1 - \frac12 + \frac13 - \cdots$ ), and the arctangent series converges at both $x = \pm 1$ .

ln(1 + x) with partial sums of its series. Inside −1 < x ≤ 1 the partial sums close in on the curve; past x = 1 they swing away.Open in grapher →

Differentiation. Differentiate $\dfrac{1}{1-x} = \sum x^n$ :

\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots = \sum_{n=1}^{\infty} n x^{n-1}, \quad |x| < 1.

Common mistake

When you integrate a series to get a specific function, find the constant of integration. For $\ln(1 + x)$ it's $0$ because $\ln 1 = 0$ ; for other functions it might not be. Also keep track of the index: after differentiating, the $n = 0$ term (a constant) disappears, so the sum usually starts at $n = 1$ .

Other operations

A few more moves round out your toolkit.

Multiplying by a power of $x$ shifts every exponent: $x^2 e^x = x^2 + x^3 + \dfrac{x^4}{2!} + \cdots$ . The radius doesn't change.
Adding or subtracting two series term by term works on the interval where both converge: $e^x + e^{-x} = 2 + x^2 + \dfrac{x^4}{12} + \cdots$ , since the odd powers cancel.
Substituting $cx$ or $x^k$ changes the radius: the series for $\dfrac{1}{1 - 4x}$ converges when $|4x| < 1$ , so its radius is $\frac14$ , not $1$ .

Approximating integrals

Some integrals, like $\int_0^1 e^{-x^2}\,dx$ or $\int_0^1 \sin(x^2)\,dx$ , have no elementary antiderivative. Replace the integrand by its series, integrate term by term, and you get a numerical series you can sum as accurately as you like. If the result is alternating, the alternating series error bound tells you how good the approximation is.

The same approach defines important functions. The antiderivative of $e^{-x^2}$ used throughout probability has no formula in terms of familiar functions, but its series, $x - \dfrac{x^3}{3} + \dfrac{x^5}{10} - \cdots$ , converges for every $x$ and can be evaluated to any accuracy.

Worked examples

Worked example: Approximating a nonelementary integral

Use the first two nonzero terms of a series to approximate $\displaystyle\int_0^1 \sin(x^2)\,dx$ , and bound the error.

Solution. Substitute $x^2$ into the sine series: $\sin(x^2) = x^2 - \dfrac{x^6}{3!} + \dfrac{x^{10}}{5!} - \cdots$ . Integrate term by term:

\int_0^1 \sin(x^2)\,dx = \frac{1}{3} - \frac{1}{7\cdot 6} + \frac{1}{11\cdot 120} - \cdots = \frac13 - \frac{1}{42} + \frac{1}{1320} - \cdots.

The first two terms give $\dfrac13 - \dfrac1{42} = \dfrac{13}{42} \approx 0.3095$ . The series alternates with decreasing terms, so the error is at most the next term, $\dfrac{1}{1320} \approx 0.00076$ .

Worked example: Summing a series by differentiating

Find the exact sum of $\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n}$ .

Solution. From above, $\displaystyle\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$ , so multiplying by $x$ gives $\displaystyle\sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2}$ for $|x| < 1$ . At $x = \frac13$ :

\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}.

Worked example: A limit using series

Find $\displaystyle\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}$ .

Solution. Replace $e^x$ with its series:

\frac{e^x - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \cdots}{x^2} = \frac12 + \frac{x}{6} + \cdots \;\longrightarrow\; \frac12.

Worked example: A function defined by a series

Let $f(x) = \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n\cdot 2^n}$ . Find the radius of convergence of the series for $f'(x)$ , and write $f'(x)$ as a familiar function.

Solution. Differentiating term by term, $f'(x) = \displaystyle\sum_{n=1}^{\infty} \frac{x^{n-1}}{2^n} = \frac12\sum_{n=1}^{\infty}\left(\frac{x}{2}\right)^{n-1}$ . This is geometric with ratio $\frac{x}{2}$ , so it converges for $|x| < 2$ , and $R = 2$ (the same as for $f$ ). Its sum is

f'(x) = \frac{1/2}{1 - x/2} = \frac{1}{2 - x}.

Tip

To find a series for a function, look for a way to reach it from a known series: substitute, multiply by a power of $x$ , differentiate, or integrate. Computing derivatives with the Taylor formula is a last resort.