From Finite Cancellation to an Infinite Sum
A finite telescoping sum is one in which many terms cancel, leaving only a small number of terms at the ends. An infinite telescoping series uses the same pattern, but it introduces an additional question: what happens to the terms left at the moving end as the number of terms grows?
The answer comes from the definition of an infinite series. Its value, when it exists, is the limit of its finite partial sums. Thus, cancellation helps simplify each partial sum, but a limit is still needed to determine convergence and the sum.
The Partial-Sum Definition
For a series \(\sum_{n=1}^{\infty} a_n\), the \(N\)th partial sum is \(S_N=\sum_{n=1}^{N}a_n\). The series converges to \(S\) if the sequence of partial sums has a finite limit \(S\) as \(N\) tends to infinity. In that case, \(S\) is called the sum of the series.
A common telescoping form is \(a_n=b_n-b_{n+1}\), where \(b_n\) is a sequence of numbers. Each positive term \(b_n\) is canceled by the negative term \(-b_n\) in the preceding summand, except for terms at the two ends.
The condition about the limit matters. Cancellation gives a formula for each partial sum, but it does not by itself guarantee that these partial sums approach a finite value.
Why the Boundary-Term Formula Holds
Here is a direct derivation. Write out the first several terms of the partial sum. The pattern shows which terms cancel and which remain; the derivation also verifies the formula for an arbitrary \(N\).
Using \(a_n=b_n-b_{n+1}\), write \(S_N=(b_1-b_2)+(b_2-b_3)+\cdots+(b_N-b_{N+1})\).
The terms \(-b_2\) and \(b_2\) sum to zero; the same is true for \(-b_3\) and \(b_3\), continuing through \(-b_N\) and \(b_N\).
Only \(b_1\) at the beginning and \(-b_{N+1}\) at the end remain, so \(S_N=b_1-b_{N+1}\).
If \(b_{N+1}\to L\), then \(S_N=b_1-b_{N+1}\to b_1-L\). This is the series sum by the partial-sum definition.
This argument identifies the essential feature of an infinite telescoping series: the first term is fixed, while the last uncanceled term changes with \(N\). The limit of that moving boundary term determines the outcome.
Worked Examples
Worked Example: A Geometric Difference
Evaluate \(\sum_{n=1}^{\infty}\left(\frac{1}{2^n}-\frac{1}{2^{n+1}}\right)\).
First, write the partial sum through the \(N\)th term:
Each negative term cancels the next positive term. The endpoints give
For a direct finite check, take \(N=3\). The terms give \((\frac12-\frac14)+(\frac14-\frac18)+(\frac18-\frac1{16})=\frac12-\frac1{16}=\frac7{16}\), which agrees with the formula \(S_3=\frac12-\frac1{16}\).
As \(N\to\infty\), \(1/2^{N+1}\to 0\). Thus \(S_N\to 1/2\), so the series converges to \(\frac12\).
Check: The partial-sum expression gives the stated value because its only \(N\)-dependent term, \(1/2^{N+1}\), tends to zero.
Worked Example: Differences of Reciprocal Squares
Evaluate \(\sum_{n=1}^{\infty}\left(\frac{1}{n^2+1}-\frac{1}{(n+1)^2+1}\right)\).
Write the first terms of \(S_N\), keeping the shared terms visible:
The negative fraction in each pair cancels the matching positive fraction in the next pair. Hence
For \(N=3\), the sum is \((\frac12-\frac15)+(\frac15-\frac1{10})+(\frac1{10}-\frac1{17})=\frac12-\frac1{17}=\frac{15}{34}\). The general formula gives \(S_3=\frac12-\frac1{17}=\frac{15}{34}\), confirming the cancellation and the endpoint.
As \(N\to\infty\), the denominator \((N+1)^2+1\) grows without bound, so its reciprocal tends to zero. Therefore \(S_N\to\frac12\), and the series sums to \(\frac12\).
Check: The limiting partial sum is \(1/2-0=1/2\), consistent with the finite sums, which retain the first term and subtract a positive residual approaching zero.
Worked Example: Alternating Boundary Terms
Let \(b_n=\frac{(-1)^{n+1}}{n+1}\), and evaluate the telescoping series \(\sum_{n=1}^{\infty}(b_n-b_{n+1})\).
The first values are \(b_1=\frac12\), \(b_2=-\frac13\), and \(b_3=\frac14\). Thus the first two series terms are \(b_1-b_2=\frac12-(-\frac13)=\frac56\) and \(b_2-b_3=-\frac13-\frac14=-\frac7{12}\). The partial sum through two terms is \(\frac56-\frac7{12}=\frac14\).
For an arbitrary \(N\), finite cancellation gives
At \(N=2\), the formula yields \(S_2=\frac12-\frac{(-1)^4}{4}=\frac12-\frac14=\frac14\), matching the direct calculation. Since the absolute value of the residual term is \(1/(N+2)\), it tends to zero. Therefore \(S_N\to\frac12\), and the series converges to \(\frac12\).
Check: Substituting \(N=2\) into the partial-sum formula reproduces the explicitly calculated two-term sum, and the residual tends to zero as required.
Convergence Depends on the Residual
In the examples, the term subtracted from the first term approaches zero, so the partial sums approach a finite limit. But a telescoping pattern does not always produce convergence. For instance, if \(b_n=n\), then \(b_n-b_{n+1}=n-(n+1)=-1\). The partial sums are \(S_N=1-(N+1)=-N\), which do not approach a finite limit. Cancellation still occurs in the expanded sum, but the uncanceled boundary term grows instead of settling down.
For a series in the form \(b_n-b_{n+1}\), inspect \(b_{N+1}\) after writing the partial sum. If it approaches a finite number \(L\), the series converges to \(b_1-L\). If it does not approach a finite limit, the partial sums do not approach a finite limit either, so the series does not have a finite sum.
A Common Cancellation Mistake
A frequent error is to say that every term cancels, and therefore the infinite sum is zero. This overlooks the uncanceled endpoints of the finite partial sums. Consider again the series whose terms are \(\frac{1}{2^n}-\frac{1}{2^{n+1}}\). If one informally cancels every displayed term, one might incorrectly report zero.
But cancellation in the first three terms gives
not zero. In general, \(S_N=\frac12-\frac{1}{2^{N+1}}\), and its limit is \(\frac12\), not zero. The correct procedure is to preserve the first and last surviving terms in every finite sum, then take the limit of the last one.
Why the Partial-Sum Method Matters
The partial-sum method gives more than a shortcut for finding an answer. It shows precisely why the series converges, identifies its value, and makes clear which boundary term controls the result. It also prevents a misleading use of cancellation: an infinite list cannot be simplified by deleting terms without first establishing what its finite partial sums are doing.
When you meet a telescoping series, organize the reasoning in this order:
- Write the \(N\)th partial sum using a finite number of terms.
- Cancel matching interior terms while keeping both endpoints.
- Find the limit of the resulting partial-sum formula.
- State convergence and the sum only if that limit is finite.
This approach applies whether the terms are positive, negative, or alternating. What matters is the actual limit of the partial sums, not the visual impression that many terms disappear.
Check Your Understanding
For each question, start with a finite partial sum and keep track of the surviving terms.
- If \(a_n=\frac{1}{n+2}-\frac{1}{n+3}\), what is the \(N\)th partial sum when the index starts at \(n=1\), and what is the infinite sum?
- For \(a_n=b_n-b_{n+1}\), derive the finite partial-sum formula through \(N\) terms.
- If \(b_1=4\) and \(b_{N+1}\to 2\), what is the sum of \(\sum_{n=1}^{\infty}(b_n-b_{n+1})\)?
- Why is it not enough to say that all terms cancel when evaluating an infinite telescoping series?
- For \(b_n=n\), find \(S_N\) for the series with terms \(b_n-b_{n+1}\). Does it converge to a finite sum?