Proof of Term-by-Term Integration

The Proof Strategy

In Term-by-Term Integration, we established the formula for integrating a power series and identified the two convergence arguments behind it. Here we give a proof that makes each passage from finite sums to an infinite series explicit. For an interior point, the key is uniform convergence on the closed segment from the center to that point. At a boundary point, the key is the Uniform Convergence Along a Convergent Boundary Segment Lemma: convergence of the original series at the endpoint supplies uniform control along the whole segment.

The calculation for a finite partial sum is elementary. The proof’s central task is to show that the integrals of those partial sums tend to the integral of the sum function. A quantitative estimate makes the role of uniform convergence particularly clear.

A Quantitative Estimate for Integrals

Theorem (Uniform Integral Error Bound): Suppose $g_N$ and $g$ are continuous on $[u,v]$, where $u\leq v$. Then $$ \sup_{x\in[u,v]}\left|\int_u^x g_N(t)\,dt-\int_u^x g(t)\,dt\right| \leq (v-u)\sup_{t\in[u,v]}|g_N(t)-g(t)|. $$

Proof. Fix $x\in[u,v]$. By the integral estimate,

\left|\int_u^x g_N(t)\,dt-\int_u^x g(t)\,dt\right| =\left|\int_u^x (g_N(t)-g(t))\,dt\right| \leq (x-u)\sup_{t\in[u,v]}|g_N(t)-g(t)|.

Because $x-u\leq v-u$, the right-hand side is at most $(v-u)\sup_{[u,v]}|g_N-g|$. This bound holds for every $x\in[u,v]$; taking the supremum over $x$ proves the estimate. If $g_N$ converges uniformly to $g$, the right-hand side tends to zero. Thus the integrals from $u$ to $x$ converge uniformly as functions of $x$, not merely at one fixed endpoint. $\square$

This estimate is a quantitative form of the Integration of a Uniform Limit Theorem. Its importance here is that a power series is first handled through its finite partial sums. Once their uniform error is controlled, the same control bounds the error in every integral along the segment.

Proving the Interior Formula

Theorem (Term-by-Term Integration of a Power Series): Suppose $\sum_{n=0}^{\infty}c_n(x-a)^n$ has radius of convergence $R>0$, and let $f$ be its sum for $|x-a|<R$. For every such $x$, $$ \int_a^x f(t)\,dt =\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}. $$ If the original series converges at a boundary point $b$, the same formula holds at $x=b$, with $f(b)$ defined by the convergent series there.

Proof. For each nonnegative integer $N$, define the polynomial partial sum $$ S_N(t)=\sum_{n=0}^{N}c_n(t-a)^n. $$ First fix an interior point $x\ne a$. Let $I$ be the closed segment with endpoints $a$ and $x$. For every $t\in I$, $|t-a|\leq |x-a|<R$. The Uniform Convergence Inside the Radius Theorem therefore gives uniform convergence of $S_N$ to $f$ on $I$. The functions $S_N$ and $f$ are continuous there; continuity of $f$ also follows from the Uniform Limits of Continuous Functions Are Continuous Theorem. By the Integration of a Uniform Limit Theorem,

\int_a^x f(t)\,dt =\lim_{N\to\infty}\int_a^x S_N(t)\,dt.

For each finite $N$, linearity of the integral and the power rule give

\int_a^x S_N(t)\,dt =\sum_{n=0}^{N}c_n\int_a^x(t-a)^n\,dt =\sum_{n=0}^{N}\frac{c_n}{n+1}(x-a)^{n+1}.

The power-rule evaluation is valid whether $x$ is to the left or the right of $a$, because the integral is oriented. Passing to the limit in these finite-sum identities proves the claimed series formula. When $x=a$, the integral is zero and every term on the right is zero, so the formula holds there as well.

This proof also gives a direct error-control interpretation. On $I$, the difference between the integral of $f$ and the integral of $S_N$ is bounded by the length of $I$ times $\sup_{t\in I}|f(t)-S_N(t)|$. That supremum tends to zero by uniform convergence, which is precisely what allows the limit to pass through the integral.

Proving the Boundary Case

Now let $b$ satisfy $|b-a|=R$, and suppose the original series converges at $b$. Set $d_n=c_n(b-a)^n$. Then $\sum_{n=0}^{\infty}d_n$ converges, and at points on the segment from $a$ to $b$ we can write $$ x=a+s(b-a),\qquad 0\leq s\leq1. $$ The partial sums along this parameterized segment are $$ S_N(a+s(b-a))=\sum_{n=0}^{N}d_n s^n. $$ By the Uniform Convergence Along a Convergent Boundary Segment Lemma, these partial sums converge uniformly for $0\leq s\leq1$. Their limit is the sum of the power series along the segment, including the value defined by the convergent series at $b$.

To apply the integration theorem without overlooking orientation, define $g_N(s)=S_N(a+s(b-a))$ and let $g$ be the corresponding limit. Uniform convergence on $[0,1]$ and the Integration of a Uniform Limit Theorem give $$ \int_0^1 g(s)\,ds=\lim_{N\to\infty}\int_0^1 g_N(s)\,ds. $$ The change of variables $t=a+s(b-a)$ gives $dt=(b-a)\,ds$. This substitution works for $b<a$ as well as $b>a$, with the sign carried by $b-a$. Consequently,

\int_a^b f(t)\,dt =(b-a)\int_0^1g(s)\,ds =\lim_{N\to\infty}(b-a)\int_0^1g_N(s)\,ds =\lim_{N\to\infty}\int_a^b S_N(t)\,dt.

Integrating each finite partial sum and using the power rule now yields $$ \int_a^b S_N(t)\,dt =\sum_{n=0}^{N}\frac{c_n}{n+1}(b-a)^{n+1}. $$ Taking the limit proves the term-by-term formula at $b$. The endpoint hypothesis matters: it is what makes $\sum d_n$ converge and allows the boundary-segment lemma to be applied. $\square$

Worked Examples

Worked Example: Integrating to an Absolutely Convergent Endpoint

Consider $$ F(x)=\sum_{n=0}^{\infty}\frac{x^n}{(n+1)^2}. $$ The coefficient root limit is $$ \lim_{n\to\infty}\left(\frac{1}{(n+1)^2}\right)^{1/n}=1, $$ so the radius of convergence is $1$. At $x=1$, the original series is $\sum_{n=0}^{\infty}1/(n+1)^2$, which converges by the $p$-Series Convergence Criterion with $p=2$. The boundary case of the theorem therefore applies:

\int_0^1 F(t)\,dt =\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}\int_0^1t^n\,dt =\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}.

For each $n$, the integral is $\int_0^1t^n\,dt=1/(n+1)$, so the resulting term is $1/(n+1)^3$. The original series converges at the endpoint, which is the required justification for passing from its partial sums to the integral over the full interval.

Worked Example: Checking the Sign on a Reversed-Side Segment

Let $$ H(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{\sqrt{n+1}}(x+2)^n. $$ The center is $a=-2$. The coefficient root limit is $1$, so the radius is $1$. At $b=-1$, the series becomes $\sum_{n=0}^{\infty}(-1)^n/\sqrt{n+1}$, which converges by the Alternating Series Test. Thus the boundary formula applies on the segment from $-2$ to $-1$.

Evaluate each integrated term at the endpoints:

\int_{-2}^{-1}\frac{(-1)^n}{\sqrt{n+1}}(t+2)^n\,dt =\frac{(-1)^n}{\sqrt{n+1}} \left[\frac{(t+2)^{n+1}}{n+1}\right]_{-2}^{-1} =\frac{(-1)^n}{(n+1)^{3/2}}.

Indeed, $t+2$ equals $0$ at the lower endpoint $t=-2$, and equals $1$ at the upper endpoint $t=-1$. The endpoint difference is therefore $(1^{n+1}-0^{n+1})/(n+1)=1/(n+1)$. For $n=0$, the term is the constant $1$, and the integral is $1$, in agreement with the formula. Term-by-term integration gives $$ \int_{-2}^{-1}H(t)\,dt =\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^{3/2}}. $$

Worked Example: An Interior Integral with a Finite Radius

For $|x-3|<1/2$, the geometric series gives $$ J(x)=\sum_{n=0}^{\infty}2^n(x-3)^n =\frac{1}{1-2(x-3)}. $$ Its radius of convergence about $3$ is $1/2$. For an interior point $x$, term-by-term integration yields

\int_3^x J(t)\,dt =\sum_{n=0}^{\infty}2^n\int_3^x(t-3)^n\,dt =\sum_{n=0}^{\infty}\frac{2^n}{n+1}(x-3)^{n+1}.

The identity can be checked against direct integration. Put $h=x-3$. Since $|h|<1/2$, $1-2h>0$, and $$ \int_3^x\frac{1}{1-2(t-3)}\,dt =\int_0^h\frac{ds}{1-2s}. $$ Once the natural logarithm is introduced, this integral equals $-\frac12\ln(1-2h)$, and the series above is its expansion. The finite partial sums converge uniformly on the segment from $3$ to $x$, so the theorem justifies their integration before the limit is taken.

What the Hypotheses Do—and Do Not—Provide

The proof uses uniform convergence on a particular closed segment, not an assumption that the series converges uniformly everywhere in its interval of convergence. For an interior point, the segment stays strictly inside the radius, so the Uniform Convergence Inside the Radius Theorem applies. At a boundary point, convergence of the series at that specific point supplies uniform convergence along the segment from the center. Neither argument gives permission to extend the formula past the radius.

A second point to check is orientation. When an endpoint lies to the left of the center, the integral from the center to that endpoint is still valid; the power rule and the change of variables retain the correct sign automatically. Direct endpoint substitution is a useful check: evaluate the transformed variable at both endpoints before simplifying the difference.

Takeaway: For finite partial sums, integration term by term is an exact algebraic calculation. Uniform convergence on the relevant segment then allows the limit of those integrals to equal the integral of the sum, both at interior points and at boundary points where the original series converges.

Check Your Understanding

Use the proof strategy and estimates in this tutorial to answer the following questions.

How does the Uniform Integral Error Bound convert a uniform error estimate for functions into an error estimate for their integrals?
Why does the segment from the center to an interior point lie in a region where uniform convergence is available?
What convergence condition at a boundary point lets the boundary-segment argument apply?
In an endpoint calculation, what values does $t+2$ take at $t=-2$ and $t=-1$?
Why does the change of variables from the center to a boundary point handle a boundary point on either side of the center?