Why telescoping helps at a boundary
Inside the radius of convergence, a power series converges by the usual power-series convergence results. At either boundary, however, the powers no longer shrink geometrically. The endpoint series must be examined on its own. When its coefficients have a suitable difference form, its partial sums may simplify dramatically.
A basic example is the coefficient \(1/[n(n+1)]\). The key observation is that it can be written as a difference of two reciprocal integers. At the positive boundary, this produces cancellation between consecutive terms. At the negative boundary, the terms have alternating signs, so the same cancellation pattern does not apply directly. The two endpoints therefore need separate analysis.
The basic telescoping identity
Partial fractions reveal the useful structure of the coefficient:
For a finite upper index \(N\), cancellation gives
Since \(1/(N+1)\) tends to zero, the infinite sum is \(1\). This finite calculation is more than a convenient shortcut: it proves convergence and identifies the sum, because an infinite series is defined as the limit of its partial sums.
Why the positive and negative boundaries differ
Consider the power series \(\sum_{n=1}^{\infty}x^n/[n(n+1)]\). Its coefficients are positive. The ratio of consecutive absolute terms, for nonzero \(x\), tends to \(|x|\), so its radius of convergence is \(1\). For \(|x|<1\), the series converges; for \(|x|>1\), its terms do not tend to zero. The boundary values are \(x=1\) and \(x=-1\).
At \(x=1\), the coefficient identity gives a telescoping series. At \(x=-1\), the terms instead have factors \((-1)^n\). The signed series does not telescope in the same immediate way. Nevertheless, its absolute-value series is exactly the positive telescoping series, so the negative endpoint converges absolutely as well.
Check whether the endpoint coefficients have a difference form and telescope the finite partial sums.
Keep the alternating factor visible; do not treat it as if it preserved the positive endpoint cancellation.
If the absolute-value series is the positive telescoping series and has a finite sum, the negative endpoint converges absolutely.
Worked endpoint calculations
Worked Example: Both endpoints of a basic power series
Find the interval of convergence and the sums at the endpoints of
Step 1: Find the radius. The absolute value of the ratio of consecutive terms is
Thus the series converges for \(|x|<1\), and its radius is \(1\). For \(|x|>1\), the term magnitudes \( |x|^n/[n(n+1)]\) do not tend to zero, so the series diverges. We must check \(x=1\) and \(x=-1\) separately.
Step 2: Positive endpoint. At \(x=1\), the \(N\)th partial sum is
Taking \(N\) to infinity gives \(S(1)=1\).
Step 3: Negative endpoint. At \(x=-1\), the absolute-value series is
Therefore the series at \(x=-1\) converges absolutely. To find its sum, use the standard identity \(\sum_{n=1}^{\infty}(-1)^{n-1}/n=\ln 2\). It follows, by shifting the index, that \(\sum_{n=1}^{\infty}(-1)^n/n=-\ln 2\) and \(\sum_{n=1}^{\infty}(-1)^n/(n+1)=\ln 2-1\). Hence
Check. The positive endpoint partial sums equal \(1-1/(N+1)\), whose limit is \(1\), as claimed. At the negative endpoint, the first term is \(-1/2\), and \(1-2\ln 2\) is approximately \(-0.38629\), consistent with an absolutely convergent sum that begins with a negative term. The interval of convergence is \([-1,1]\).
The identity involving \(\ln 2\) can be justified directly if needed. For \(0\leq t\leq1\), the finite geometric expansion of \(1/(1+t)\) has a remainder whose absolute value is at most \(t^N\). Integrating from \(0\) to \(1\), the remainder has absolute integral at most \(1/(N+1)\), which tends to zero. The integrals of the finite sums therefore tend to \(\int_0^1dt/(1+t)=\ln 2\), giving the alternating harmonic identity used above.
Worked Example: A rescaled series with different endpoint values
Determine the convergence and endpoint sums of
Step 1: Locate the boundary. The ratio of consecutive absolute terms tends to \(|x|/2\). Thus the radius is \(2\): the series converges for \(|x|<2\), and for \(|x|>2\) its terms do not tend to zero. The endpoints are \(x=2\) and \(x=-2\).
Step 2: Evaluate at \(x=2\). Substitution gives
Step 3: Evaluate at \(x=-2\). The endpoint series is \(\sum_{n=1}^{\infty}(-1)^n/[n(n+1)]\). Its absolute-value series sums to \(1\), so it converges absolutely. Using the shifted alternating harmonic sums calculated above, its value is \(1-2\ln2\).
Check. At \(x=2\), the finite partial sum is \(1-1/(N+1)\) and tends to \(1\). At \(x=-2\), the terms are exactly those at \(x=-1\) in the first example, so the endpoint sum \(1-2\ln2\) agrees. The interval of convergence is \([-2,2]\).
Worked Example: A shifted telescoping coefficient
Analyze the endpoint behavior of
Step 1: Find the radius. The ratio of consecutive absolute terms tends to \(|x|/4\), so the radius is \(4\). For \(|x|>4\), the terms do not tend to zero. The boundary points are \(x=4\) and \(x=-4\).
Step 2: Rewrite the coefficient.
At \(x=4\), the partial sum becomes
Letting \(N\) tend to infinity gives \(S(4)=3/4\).
Step 3: Check \(x=-4\). The absolute-value series at this endpoint is the positive endpoint series, with sum \(3/4\). Thus the negative endpoint converges absolutely. Its sum can also be evaluated:
Here the shifted sum in parentheses follows by removing the first two terms from \(\sum_{k=1}^{\infty}(-1)^k/k=-\ln2\): the omitted terms sum to \(-1+1/2=-1/2\), so the remaining tail is \(1/2-\ln2\).
Check. The positive partial sums tend to \(3/4\). At the negative endpoint, the first term is \(-1/3\), and the exact sum \(-1/4\) has the expected negative sign. Absolute convergence is confirmed because the corresponding positive series sums to \(3/4\). The interval of convergence is \([-4,4]\).
A useful general justification
The positive endpoint argument extends beyond the particular coefficient \(1/[n(n+1)]\). Suppose \(a_n\geq0\), \(a_n\) decreases, and \(a_n\) tends to zero. Then the positive series with terms \(a_n-a_{n+1}\) has partial sums \(a_1-a_{N+1}\), so its sum is \(a_1\). At the negative boundary, the terms become \((-1)^n(a_n-a_{n+1})\). They may no longer telescope, but their absolute values still sum to \(a_1\). Therefore that endpoint is absolutely convergent too.
The condition matters: the signed endpoint sum generally is not \(a_1\). The cancellation that gives \(a_1\) applies to the positive telescoping series, not automatically to its alternating version. Finding the negative endpoint’s exact value may require a separate calculation, as the examples illustrate.
A common mistake
A tempting but incorrect step is to treat the negative endpoint as though the positive telescoping cancellation still gave the same sum. For instance, one might write
This drops the factor \((-1)^n\), changing the series. The correct substitution is
and its sum is \(1-2\ln2\), not \(1\). The right conclusion from the positive telescoping calculation at the negative endpoint is that the series of absolute values converges; it does not determine the signed sum by itself.
Why the method is useful
Telescoping exposes endpoint behavior that may be hard to see from the original coefficient. A decomposition into a difference can give an exact positive-boundary sum in a few lines, and the same calculation often proves absolute convergence at the negative boundary. It also separates two questions that should never be conflated: whether an endpoint series converges, and what its sum is.
For a power series, the procedure is dependable: find the radius using the usual power-series tests, substitute each boundary value separately, and then use the coefficient structure at that particular endpoint. At the positive boundary, telescope the partial sums. At the negative boundary, first check absolute convergence; only then calculate its signed sum if an exact value is needed.
Check Your Understanding
For each question, keep the finite partial sums and the endpoint substitutions explicit.
- Rewrite \(1/[n(n+1)]\) as a difference and calculate \(\sum_{n=1}^{N}1/[n(n+1)]\).
- For \(\sum_{n=1}^{\infty}x^n/[n(n+1)]\), what are the boundary points, and why must they be checked separately?
- Why does convergence of \(\sum_{n=1}^{\infty}1/[n(n+1)]\) imply absolute convergence at the negative boundary?
- Explain the error in replacing \(\sum(-1)^n(1/n-1/(n+1))\) by \(\sum(1/n-1/(n+1))\).
- Rewrite \(1/[n(n+2)]\) as a difference and find the sum of its positive endpoint series.
