How Integration Changes a Power Series
Integrating a power series term by term changes its coefficients. That change can be especially important at the endpoints of the interval of convergence, where even a small improvement in the size of the terms can determine whether the resulting series converges.
Consider the series with general term \(x^n/n^p\), where \(n\) ranges over the positive integers and \(p\) is a fixed real number. Its term-by-term integral from \(0\) to \(x\) has terms \(x^{n+1}/((n+1)n^p)\). The new factor \(1/(n+1)\) is the feature we will analyze.
The lower limit \(0\) fixes the constant of integration. For a general antiderivative, an arbitrary constant could be added.
Why Term-by-Term Integration Works Inside the Interval
The displayed integration rule is valid for \(|x|<1\). Here is the justification. A power series converges uniformly on every closed interval strictly inside its radius of convergence. In particular, for any fixed \(x\) with \(|x|<1\), the partial sums \(\sum_{n=1}^N t^n/n^p\) converge uniformly on the interval between \(0\) and \(x\).
Uniform convergence allows the integral of the limit to be obtained as the limit of the integrals. To see the key estimate, write \(S_N(t)=\sum_{n=1}^N t^n/n^p\). If \(f(t)\) is the sum, then \[ \left|\int_0^x f(t)\,dt-\int_0^x S_N(t)\,dt\right| \leq |x|\sup_{t\text{ between }0\text{ and }x}|f(t)-S_N(t)|. \] The supremum on the right tends to zero by uniform convergence. Meanwhile, integrating the finite sum gives \[ \int_0^x S_N(t)\,dt =\sum_{n=1}^N\frac{x^{n+1}}{(n+1)n^p}. \] Taking the limit proves the term-by-term formula for \(|x|<1\).
This argument applies inside the interval. It does not by itself establish that the integrated series converges at \(x=1\) or \(x=-1\). Each endpoint must be tested separately.
The Coefficient Change at the Endpoints
At \(x=1\), the integrated series is positive: \[ \sum_{n=1}^{\infty}\frac{1}{(n+1)n^p}. \] At \(x=-1\), its terms alternate in sign: \[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(n+1)n^p}. \] The magnitude of the terms at either endpoint is \(b_n=1/((n+1)n^p)\).
For every real \(p\), compare this magnitude with \(1/n^{p+1}\). The ratio is \[ \frac{b_n}{1/n^{p+1}} =\frac{n}{n+1} \longrightarrow 1. \] Thus the positive series of endpoint magnitudes has the same convergence behavior as the \(p\)-series with exponent \(p+1\). That series converges when \(p+1>1\), or \(p>0\), and diverges when \(p\leq 0\).
- If \(p>0\), the series converges absolutely at both \(x=1\) and \(x=-1\).
- If \(-1<p\leq 0\), it diverges at \(x=1\) and converges conditionally at \(x=-1\).
- If \(p\leq -1\), it diverges at both endpoints.
To complete the reasoning for \(x=-1\), suppose \(p>-1\). Then \(b_n\) tends to zero because it behaves like \(1/n^{p+1}\), whose exponent is positive. The terms also decrease eventually. For example, the function \(g(t)=t^{-p}/(t+1)\) has derivative \[ g'(t)=\frac{-(p+1)t-p}{t^{p+1}(t+1)^2}. \] When \(p>-1\), the numerator is negative for all sufficiently large \(t\), so \(g(t)\), and hence \(b_n\), eventually decreases. The alternating-series test therefore gives convergence at \(-1\). If \(-1<p\leq0\), the series of magnitudes diverges, so this convergence is conditional. If \(p>0\), the series of magnitudes converges, giving absolute convergence.
For \(p\leq -1\), the endpoint terms do not tend to zero: \(b_n\) behaves like \(n^{-(p+1)}\), with exponent at most zero. The term test rules out convergence at either endpoint. At \(x=1\), the positive series also diverges whenever \(p\leq0\), as the comparison with the \(p\)-series shows.
Worked Examples
Worked Example: Integrating the Series with \(p=2\)
Start with \(\sum_{n=1}^{\infty}x^n/n^2\). Integrate each term from \(0\) to \(x\):
At \(x=1\), the terms are positive, and their magnitudes can be compared with \(1/n^3\):
Since \(\sum 1/n^3\) converges, the integrated series converges at \(1\). At \(x=-1\), the terms have the same magnitudes, so the series converges absolutely there as well.
Check the integration inside \(|x|<1\): differentiating the integrated term gives \[ \frac{d}{dx}\left(\frac{x^{n+1}}{(n+1)n^2}\right) =\frac{(n+1)x^n}{(n+1)n^2} =\frac{x^n}{n^2}, \] which recovers each original term.
Worked Example: The Borderline Case \(p=0\)
When \(p=0\), the original terms are \(x^n\). Integrating them gives \[ \int_0^x t^n\,dt =\left[\frac{t^{n+1}}{n+1}\right]_0^x =\frac{x^{n+1}}{n+1}. \] At \(x=1\), the resulting series is \(\sum_{n=1}^{\infty}1/(n+1)\). Its terms are the harmonic terms starting at \(1/2\), so the series diverges.
At \(x=-1\), the series is \[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}. \] Its magnitudes \(1/(n+1)\) decrease to zero, so the alternating-series test gives convergence. The series of magnitudes diverges, so the convergence is conditional.
Check by differentiating a term: \[ \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) =\frac{(n+1)x^n}{n+1} =x^n. \] The integrated term differentiates back to the original term.
Worked Example: A Conditional Endpoint with \(p=-\tfrac12\)
Here \(n^p=\sqrt{n}^{\,-1}=1/\sqrt n\), so the original term is \(\sqrt n\,x^n\). Its integral from \(0\) to \(x\) is \[ \int_0^x \sqrt n\,t^n\,dt =\sqrt n\left[\frac{t^{n+1}}{n+1}\right]_0^x =\frac{\sqrt n\,x^{n+1}}{n+1}. \] At \(x=1\), the positive terms are \(\sqrt n/(n+1)\). Their ratio to \(1/\sqrt n\) is \[ \frac{\sqrt n/(n+1)}{1/\sqrt n} =\frac{n}{n+1} \longrightarrow 1. \] The series \(\sum 1/\sqrt n\) diverges, so the integrated series diverges at \(1\).
At \(x=-1\), the terms alternate and have magnitudes \(b_n=\sqrt n/(n+1)\). These tend to zero, since \(b_n\) behaves like \(1/\sqrt n\). To check eventual decrease, let \(g(t)=\sqrt t/(t+1)\). Then \[ g'(t)=\frac{1-t}{2\sqrt t\,(t+1)^2}<0 \quad\text{for }t>1. \] The alternating-series test gives convergence at \(-1\). Because the magnitudes form a divergent series, the convergence is conditional.
Finally, differentiating an integrated term verifies the coefficient: \[ \frac{d}{dx}\left(\frac{\sqrt n\,x^{n+1}}{n+1}\right) =\frac{\sqrt n(n+1)x^n}{n+1} =\sqrt n\,x^n. \]
A Common Endpoint Mistake
A frequent mistake is to assume that the extra denominator \(n+1\) guarantees convergence. For \(p=0\), the integrated terms at \(x=1\) are \(1/(n+1)\). It is incorrect to conclude that their series converges just because the denominator has increased.
The correction is to compare the whole coefficient with a known series. Integration changes the approximate size from \(n^{-p}\) to \(n^{-(p+1)}\). That shift improves the convergence threshold, but it does not make every endpoint series converge.
Why the Endpoint Test Matters
Inside \(|x|<1\), the integrated series is justified by the power-series integration argument. At the boundary, however, the factor \(x^{n+1}\) no longer decreases in magnitude. The coefficient \(1/((n+1)n^p)\) then determines absolute convergence, while the alternating sign at \(x=-1\) can provide convergence even when the magnitudes do not sum to a finite value.
The threshold shift is worth remembering. At \(x=1\), the integrated series converges for \(p>0\), whereas the corresponding positive series \(\sum 1/n^p\) requires \(p>1\). At \(x=-1\), alternation permits conditional convergence for the additional range \(-1<p\leq0\). These conclusions follow from testing the integrated series itself at each boundary, not from extending an interior integration argument to the endpoints.
Check Your Understanding
Use the factor \(1/(n+1)\) and test the endpoints separately.
- Write the term-by-term integral from \(0\) to \(x\) of \(\sum_{n=1}^{\infty}x^n/n^p\).
- For \(p=1/2\), determine whether the integrated series converges absolutely at each endpoint.
- For \(p=0\), explain why the integrated series diverges at \(x=1\) but converges conditionally at \(x=-1\).
- For \(p=-1\), what happens to the magnitude of the integrated term at either endpoint, and what does the term test imply?
- Why does convergence of the integrated series inside \((-1,1)\) not by itself settle convergence at \(x=1\) or \(x=-1\)?
