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Section 6.2 Calculus of Power Series

Inside the interval of convergence, power series behave much like polynomials. In particular, they can be differentiated and integrated term by term. The full proof uses ideas about uniform convergence, but the behavior of the new coefficients is already visible from the root test.

Subsection 6.2.1

Proof.
Let
\begin{equation*} L=\limsup |a_n|^{1/n}, \end{equation*}
so that \(R=1/L\) by Theorem 6.1.2. Since \(n^{1/n}\to 1\) by Example 2.2.7, we have
\begin{equation*} \limsup |na_n|^{1/n} = \limsup n^{1/n}|a_n|^{1/n} = L. \end{equation*}
Hence the formal derivative has radius \(R\text{.}\)
Likewise, because \((n+1)^{1/n}\to 1\text{,}\)
\begin{equation*} \limsup \left|\frac{a_n}{n+1}\right|^{1/n} = \limsup \frac{|a_n|^{1/n}}{(n+1)^{1/n}} = L, \end{equation*}
so the formal integral also has radius \(R\text{.}\)

Proof.
Fix \(r\) with \(0 \lt r \lt R\text{,}\) and let \(I_r=[x_0-r,x_0+r]\text{.}\) By Theorem 8.1.8, the original power series and its formal derivative both converge uniformly on \(I_r\text{.}\) Let
\begin{equation*} s_n(x)=\sum_{k=0}^n a_k(x-x_0)^k. \end{equation*}
Each \(s_n\) is a polynomial, so
\begin{equation*} s_n'(x)=\sum_{k=1}^n ka_k(x-x_0)^{k-1}. \end{equation*}
Also \(s_n(x_0)=a_0\) for every \(n\text{.}\) Therefore Theorem 8.2.8 applies and shows that the limit function \(f\) is differentiable on \((x_0-r,x_0+r)\text{,}\) with
\begin{equation*} f'(x)=\sum_{k=1}^\infty ka_k(x-x_0)^{k-1} \qquad \text{for } x \in (x_0-r,x_0+r). \end{equation*}
Likewise, since \(s_n \to f\) uniformly on \(I_r\text{,}\) Theorem 8.2.6 gives
\begin{equation*} \begin{aligned} \int_{x_0}^x f(t)\,dt &= \lim_{n\to\infty} \int_{x_0}^x s_n(t)\,dt \\ &= \lim_{n\to\infty} \sum_{k=0}^n \frac{a_k}{k+1}(x-x_0)^{k+1} \\ &= \sum_{k=0}^\infty \frac{a_k}{k+1}(x-x_0)^{k+1} \end{aligned} \end{equation*}
for every \(x \in (x_0-r,x_0+r)\text{.}\)
Since \(r \lt R\) was arbitrary, both formulas hold throughout the entire interval \((x_0-R,x_0+R)\text{.}\)

Example 6.2.3.

We can compute the sum
\begin{equation*} \sum_{n=1}^\infty \frac{n}{2^n}. \end{equation*}
By the geometric series formula (Proposition 3.2.1),
\begin{equation*} \frac{1}{1-x}=\sum_{n=0}^\infty x^n \qquad \text{for } |x|<1. \end{equation*}
Differentiating term by term gives
\begin{equation*} \frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}. \end{equation*}
Multiplying by \(x\text{,}\) we obtain
\begin{equation*} \frac{x}{(1-x)^2} = \sum_{n=1}^\infty nx^n. \end{equation*}
Evaluating at \(x=1/2\) gives
\begin{equation*} \sum_{n=1}^\infty \frac{n}{2^n} = \frac{\frac12}{(1-\frac12)^2} = 2. \end{equation*}

Example 6.2.4.

The geometric series also yields familiar expansions. Replacing \(x\) by \(-x\text{,}\) we get
\begin{equation*} \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^n x^n \qquad \text{for } |x|<1. \end{equation*}
Integrating from \(0\) to \(x\) gives
\begin{equation*} \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n \qquad \text{for } |x|<1. \end{equation*}
Likewise,
\begin{equation*} \frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^n x^{2n} \qquad \text{for } |x|<1, \end{equation*}
so integrating from \(0\) to \(x\) yields
\begin{equation*} \arctan x = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} \qquad \text{for } |x|<1. \end{equation*}

Proof.
Replacing \(a_0\) by \(a_0-L\text{,}\) we may assume without loss of generality that \(L=0\text{.}\) Let
\begin{equation*} s_k:=\sum_{n=0}^k a_n. \end{equation*}
Then \(s_k \to 0\text{.}\)
For \(|x|<1\text{,}\) the geometric series \(\sum_{m=0}^\infty x^m\) converges absolutely by Proposition 3.2.1. Hence Mertens’ theorem (Theorem 3.3.1) applies to the product of \(\sum_{m=0}^\infty x^m\) and \(\sum_{n=0}^\infty a_nx^n\text{,}\) and gives
\begin{equation*} \frac{f(x)}{1-x} = \left(\sum_{m=0}^\infty x^m\right)\left(\sum_{n=0}^\infty a_nx^n\right) = \sum_{k=0}^\infty s_kx^k. \end{equation*}
Therefore
\begin{equation*} f(x)=(1-x)\sum_{k=0}^\infty s_kx^k \qquad \text{for } 0<x<1. \end{equation*}
Let \(\varepsilon>0\text{.}\) Since \(s_k \to 0\text{,}\) there exists \(N\) such that \(|s_k|<\varepsilon\) for all \(k \ge N\text{.}\) Then for \(0<x<1\text{,}\)
\begin{equation*} \left|(1-x)\sum_{k=N}^\infty s_kx^k\right| \le (1-x)\sum_{k=N}^\infty |s_k|x^k \lt \varepsilon(1-x)\sum_{k=N}^\infty x^k = \varepsilon x^N \le \varepsilon. \end{equation*}
For the remaining finitely many terms, let
\begin{equation*} B:=\max\{|s_0|,|s_1|,\dots,|s_{N-1}|\}. \end{equation*}
Then
\begin{equation*} \left|(1-x)\sum_{k=0}^{N-1} s_kx^k\right| \le B(1-x)\sum_{k=0}^{N-1} x^k = B(1-x^N). \end{equation*}
Since \(1-x^N \to 0\) as \(x \to 1^-\text{,}\) there exists \(\delta>0\) such that \(0<1-x<\delta\) implies \(B(1-x^N)<\varepsilon\text{.}\)
Combining the two estimates, we obtain \(|f(x)|<2\varepsilon\) whenever \(0<1-x<\delta\text{.}\) Thus \(\lim_{x\to 1^-} f(x)=0\text{.}\) Undoing the initial reduction proves that \(\lim_{x\to 1^-} f(x)=L\text{.}\)
By a simple change of variables, Abel’s theorem also applies at either endpoint of a finite interval of convergence.

Example 6.2.6.

From Example 6.2.4, we know that
\begin{equation*} \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n \qquad \text{for } |x|<1. \end{equation*}
The series at \(x=1\) is the alternating harmonic series, which converges by Proposition 3.2.15. Since \(\ln(1+x)\) is continuous at \(x=1\text{,}\) Abel’s theorem gives
\begin{equation*} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \lim_{x\to 1^-}\ln(1+x) = \ln 2. \end{equation*}
These examples illustrate a general principle: once a single power series is known, algebraic manipulations together with differentiation and integration often produce many others.
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