Why Endpoints Need Their Own Tests
A power series can behave one way at points strictly inside its interval of convergence and differently at the two endpoints. Termwise differentiation and integration give new power series that agree with the corresponding calculus operations inside the interval, but that fact does not settle whether either transformed series converges at an endpoint.
The endpoint question is concrete: substitute each endpoint into the original series, then into the differentiated series, and then into the integrated series. The resulting numerical series may have different convergence behavior. In particular, differentiation can turn a convergent endpoint series into a divergent one, while integration can make some divergent endpoint series converge.
What to Test at Each Endpoint
Consider a power series centered at \(a\), with a positive finite radius \(R\):
The endpoint candidates are \(x=a+R\) and \(x=a-R\). Inside the interval \(|x-a|<R\), the differentiated and integrated series are obtained term by term. The differentiated series is
and an antiderivative series is
where \(C\) is a constant. At the endpoints, substitute \(x=a+R\) and \(x=a-R\) into each expression. The constant \(C\) has no effect on convergence.
There is a useful way to compare the three endpoint tests. Choose one endpoint and write it as \(x=a+sR\), where \(s\) is either \(1\) or \(-1\). For \(n\geq 0\), define the original endpoint term \(b_n=c_n(sR)^n\). The original series at that endpoint is \(\sum b_n\). For the derivative, the \(n\)th term, with \(n\geq 1\), becomes \(\frac{n}{sR}b_n\). For the integral, its \(n\)th term becomes \(\frac{sR}{n+1}b_n\).
Since \(sR\) is a nonzero constant, multiplying a series by \(1/(sR)\) or \(sR\) does not change whether it converges. Thus the endpoint comparison is between \(\sum b_n\), \(\sum n b_n\), and \(\sum \frac{b_n}{n+1}\). These are different series and need not have the same outcome.
A Reliable Endpoint Procedure
For a series centered at \(a\) with endpoint distance \(R\), substitute \(a-R\) and \(a+R\) separately.
Do this for the original series and for its termwise derivative or integral. Do not infer endpoint behavior from the interior.
Check, as appropriate, whether terms tend to zero, and use tests such as the geometric, \(p\)-series, alternating-series, or comparison test.
Convergence at one endpoint says nothing by itself about convergence at the other endpoint or after the other calculus operation.
Worked Example: Differentiation Loses Convergence at One Endpoint
Worked Example: Differentiating a Series with \(1/n^2\) Coefficients
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n^2}\) at the endpoints \(x=1\) and \(x=-1\). First test the original series.
At \(x=1\), substitution gives \(\sum_{n=1}^{\infty}\frac{1}{n^2}\), which converges by the \(p\)-series test because \(p=2>1\). At \(x=-1\), it gives \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\). Its absolute-value series is \(\sum 1/n^2\), so it converges absolutely.
Now differentiate each term. Since \(\frac{d}{dx}(x^n/n^2)=x^{n-1}/n\), the differentiated series is \(\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}\).
At \(x=1\), this becomes \(\sum_{n=1}^{\infty}1/n\), the harmonic series, which diverges. At \(x=-1\), it becomes \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). The positive terms \(1/n\) decrease to zero, so the alternating-series test gives convergence.
Finally, integrate each term. An antiderivative series is \(C+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n^2(n+1)}\). At either endpoint, the absolute value of the \(n\)th term is \(1/[n^2(n+1)]\), which is at most \(1/n^3\). Since \(\sum 1/n^3\) converges, the integrated series converges absolutely at both endpoints.
Check: Differentiating \(x^{n+1}/[n^2(n+1)]\) gives \(x^n/n^2\), as required. Differentiating \(x^n/n^2\) gives \(x^{n-1}/n\), also as used. Thus the endpoint series were formed from the correct termwise operations.
Why Integration Preserves Convergence When the Original Endpoint Series Converges
There is a helpful one-way conclusion for integration. If the original series converges at an endpoint, then the corresponding integrated series converges there. This does not mean that the integrated series must converge whenever the original one diverges.
To justify the first claim, use the endpoint terms \(b_n\) above. Suppose \(\sum_{n=1}^{\infty}b_n\) converges, and let \(B_n=\sum_{k=1}^{n}b_k\). The sequence \(B_n\) converges, so it is bounded: there is a constant \(M\) such that \(|B_n|\leq M\) for every \(n\). Set \(w_n=1/(n+1)\). These weights decrease to zero. Summation by parts gives
The first term tends to zero because \(B_N\) is bounded and \(w_N\to 0\). The remaining series converges absolutely: its sum of absolute values is at most \(M\sum_{n=1}^{\infty}(w_n-w_{n+1})\), and this telescoping sum is \(w_1\), a finite number. Therefore \(\sum b_n/(n+1)\) converges. Multiplying by the endpoint’s constant factor \(sR\) does not change convergence, so the integrated endpoint series converges as claimed.
Worked Example: Integration Can Change Endpoint Convergence
Worked Example: The Series \(\sum x^n/n\)
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n}\) at \(x=1\) and \(x=-1\). At \(x=1\), the original series is \(\sum 1/n\), which diverges. At \(x=-1\), it is \(\sum (-1)^n/n\). The terms \(1/n\) decrease to zero, so this alternating series converges.
Termwise differentiation gives \(\sum_{n=1}^{\infty}x^{n-1}\), because \(\frac{d}{dx}(x^n/n)=x^{n-1}\). At \(x=1\), the endpoint series is \(\sum 1\), whose terms do not tend to zero, so it diverges. At \(x=-1\), it is \(\sum (-1)^{n-1}\); its terms also do not tend to zero, so it diverges.
Termwise integration gives \(C+\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}\). At \(x=1\), the nonconstant part is \(\sum 1/[n(n+1)]\). Since \(1/[n(n+1)]=1/n-1/(n+1)\), its partial sums are \(1-1/(N+1)\), which tend to \(1\). It converges.
At \(x=-1\), the integrated series is \(\sum (-1)^{n+1}/[n(n+1)]\). It converges absolutely because the sum of the absolute values is the same telescoping series, with sum \(1\).
Check: The derivative of \(x^n/n\) is \(x^{n-1}\), and the derivative of \(x^{n+1}/[n(n+1)]\) is \(x^n/n\). At the two endpoints, the original series therefore has outcomes “diverges, converges,” the derivative “diverges, diverges,” and the integral “converges, converges.”
Worked Example: Integration Does Not Always Fix Divergence
Worked Example: The Geometric Series at \(x=1\) and \(x=-1\)
Start with the geometric power series \(\sum_{n=0}^{\infty}x^n\). At \(x=1\), the original series is \(\sum 1\), which diverges because its terms do not approach zero. At \(x=-1\), it is \(\sum (-1)^n\), which also diverges because its terms do not approach zero.
Its termwise derivative is \(\sum_{n=1}^{\infty}n x^{n-1}\). At \(x=1\), this becomes \(\sum n\), whose terms do not approach zero. At \(x=-1\), it becomes \(\sum n(-1)^{n-1}\); the terms have magnitudes \(n\), so they do not approach zero. Both endpoint derivative series diverge.
Its termwise integral, apart from a constant, is \(\sum_{n=0}^{\infty}x^{n+1}/(n+1)\). At \(x=1\), this is the harmonic series \(\sum_{n=0}^{\infty}1/(n+1)\), which diverges. At \(x=-1\), it is \(\sum_{n=0}^{\infty}(-1)^{n+1}/(n+1)\). Its alternating terms decrease in magnitude to zero, so it converges.
Check: Differentiating \(x^{n+1}/(n+1)\) gives \(x^n\), and differentiating \(x^n\) gives \(nx^{n-1}\). Integration makes the series converge at \(-1\), but not at \(1\); thus it does not automatically cure endpoint divergence.
Common Mistake: Reusing the Original Endpoint Result
A tempting but incorrect argument is: “The original series converges at \(x=-1\), so its derivative series converges there too.” The example \(\sum x^n/n\) disproves this claim. At \(x=-1\), the original series \(\sum (-1)^n/n\) converges, but the derivative series becomes \(\sum (-1)^{n-1}\). Its terms do not approach zero, so it diverges.
The correction is to substitute the endpoint into the transformed series and test that numerical series. Convergence of the original endpoint series does guarantee convergence of the integrated endpoint series, as shown above, but it gives no such guarantee for differentiation. Even when the differentiated series converges at an endpoint, that convergence alone should not be confused with a claim that the represented function has an endpoint derivative there.
Endpoint Results at a Glance
| Series being tested | Endpoint terms, up to a nonzero constant factor | What to conclude |
|---|---|---|
| Original power series | \(b_n\) | Test \(\sum b_n\) directly. |
| Termwise derivative | \(n b_n\) | Test \(\sum n b_n\) directly; convergence may be lost. |
| Termwise integral | \(b_n/(n+1)\) | Test directly; convergence follows if \(\sum b_n\) converges, but can also occur when the original series diverges. |
For each endpoint, these tests are independent. The right endpoint may behave differently from the left, and differentiation may behave differently from integration. A clear record should state the result for the original, derivative, and integral series separately at both endpoints.
Check Your Understanding
For each question, write the numerical endpoint series before deciding whether it converges.
- For \(\sum_{n=1}^{\infty}x^n/n^3\), what are the original endpoint series at \(x=1\) and \(x=-1\)?
- For \(\sum_{n=1}^{\infty}x^n/n^2\), what derivative series results at each endpoint, and which convergence tests apply?
- Why does convergence of an original endpoint series imply convergence of the corresponding integrated endpoint series?
- For \(\sum_{n=0}^{\infty}x^n\), does the integrated series converge at \(x=1\)? Explain using the endpoint series.
- Why is it invalid to infer endpoint convergence of a differentiated series from convergence of the original series at that endpoint?
