Convergence From the Boundary Terms of a Telescoping Sum

Why the Uncancelled Terms Matter

A telescoping series is useful because most terms disappear when its partial sums are expanded. The terms that remain are the boundary terms: the first term at one end and a term near the far end. As the number of summands grows, the behavior of that far-end term determines whether the partial sums settle to a finite value.

The key is to examine finite partial sums before making a claim about an infinite series. Cancellation in a finite sum is exact; convergence is then decided by the limit of the boundary sequence.

Central idea: If the terms have the form $a_n=u_n-u_{n+1}$, then the $N$th partial sum is $u_1-u_{N+1}$. The series converges to a finite sum exactly when $u_{N+1}$ has a finite limit.

Telescoping Terms and Partial Sums

A series $\sum_{n=1}^{\infty}a_n$ is defined through its partial sums $S_N=\sum_{n=1}^{N}a_n$. It converges if the sequence of partial sums approaches a finite real number as $N$ tends to infinity. The series diverges if those partial sums have no finite limit.

For a telescoping series, each summand is a difference between consecutive values of another sequence. One common arrangement is $a_n=u_n-u_{n+1}$. The opposite arrangement, $a_n=u_{n+1}-u_n$, works in the same way, with the signs reversed.

Definition: A series is telescoping when its terms can be written as differences of consecutive values of a sequence. For $a_n=u_n-u_{n+1}$, its $N$th partial sum is $S_N=u_1-u_{N+1}$. For $a_n=u_{n+1}-u_n$, it is $S_N=u_{N+1}-u_1$.

The boundary sequence is the sequence of values that remains at the far end of these finite sums: $u_{N+1}$. It need not tend to zero. It needs to approach some finite real number. If its limit is $L$, then the series sum is determined by the fixed initial boundary value and $L$.

The Finite Identity and the Convergence Criterion

For $a_n=u_n-u_{n+1}$, expand the first few terms of a partial sum:

S_N=(u_1-u_2)+(u_2-u_3)+\cdots+(u_N-u_{N+1}).

Every term $u_2,\ldots,u_N$ appears once with a positive sign and once with a negative sign, so they cancel. Only the two boundary terms remain. This finite identity gives a complete convergence test.

Theorem: Suppose $a_n=u_n-u_{n+1}$ for every $n\in\mathbb{N}$. Then $\sum_{n=1}^{\infty}a_n$ converges to a finite sum if and only if $u_n$ has a finite limit $L$. In that case, $$ \sum_{n=1}^{\infty}a_n=u_1-L. $$ If instead $a_n=u_{n+1}-u_n$, the series converges if and only if $u_n$ has a finite limit $L$, and its sum is $L-u_1$.

Here is the justification for the first arrangement. The finite identity gives $S_N=u_1-u_{N+1}$. If $u_n\to L$, then $u_{N+1}\to L$, so $S_N\to u_1-L$, a finite number. Conversely, suppose $S_N\to S$, a finite number. Rearranging the same identity gives $u_{N+1}=u_1-S_N$. Therefore $u_{N+1}\to u_1-S$, so the boundary sequence has a finite limit. This proves both directions, and taking the limit in the identity gives the sum.

For the reverse arrangement, $S_N=u_{N+1}-u_1$. The same reasoning shows that a finite limit of $u_n$ gives a finite limit of $S_N$, and that convergence of $S_N$ forces $u_{N+1}=S_N+u_1$ to have a finite limit.

Worked Examples

Worked Example: A Boundary Sequence Approaching Zero

Determine the convergence and sum of $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$.

First split the fraction into consecutive reciprocals:

\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1},

because $\frac{1}{n}-\frac{1}{n+1}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}$. Thus $u_n=1/n$. The finite partial sum is

S_N = \left(1-\frac12\right) +\left(\frac12-\frac13\right) +\cdots +\left(\frac1N-\frac{1}{N+1}\right) = 1-\frac{1}{N+1}.

The boundary value $u_{N+1}=1/(N+1)$ tends to zero. Therefore $S_N\to 1$, and the series converges to $1$.

Check: The finite formula gives $S_1=1-\frac12=\frac12$, which matches the first summand $\frac{1}{1\cdot2}=\frac12$. Taking the limit in the formula gives the stated infinite sum, $1$.

Worked Example: A Boundary Sequence That Grows Without Bound

Determine whether $\sum_{n=1}^{\infty}\bigl(\sqrt{n+1}-\sqrt n\bigr)$ converges.

These terms have the form $u_{n+1}-u_n$ with $u_n=\sqrt n$. Adding the first $N$ terms gives

S_N = (\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{N+1}-\sqrt N) = \sqrt{N+1}-1.

Since $\sqrt{N+1}$ grows without bound, $S_N$ grows without bound. The series does not converge to a finite sum; it diverges.

In fact, its individual terms do tend to zero. Rationalizing shows this directly:

\sqrt{n+1}-\sqrt n = \frac{(\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n)} {\sqrt{n+1}+\sqrt n} = \frac{1}{\sqrt{n+1}+\sqrt n} \longrightarrow 0.

This does not change the conclusion: the partial sums grow without bound. Check: At $N=2$, the original partial sum is $(\sqrt2-1)+(\sqrt3-\sqrt2)=\sqrt3-1$, exactly as the finite formula gives.

Worked Example: An Oscillating Boundary Sequence

Let $a_n=(-1)^n-(-1)^{n+1}$. Decide whether $\sum_{n=1}^{\infty}a_n$ converges.

Set $u_n=(-1)^n$, so $a_n=u_n-u_{n+1}$. The finite identity gives

S_N=u_1-u_{N+1}=-1-(-1)^{N+1}.

If $N$ is even, then $(-1)^{N+1}=-1$, so $S_N=0$. If $N$ is odd, then $(-1)^{N+1}=1$, so $S_N=-2$. The partial sums alternate between $0$ and $-2$, so they have no limit and the series diverges.

Check: The first two terms are $a_1=-2$ and $a_2=2$. Thus $S_1=-2$ and $S_2=0$, agreeing with the two cases of the formula.

Reading the Boundary Behavior

The criterion is about a finite limit, not specifically a zero limit. If $u_n\to L$, where $L$ is nonzero, the difference series can still converge. Its sum is $u_1-L$ for terms $u_n-u_{n+1}$, or $L-u_1$ for terms $u_{n+1}-u_n$. A nonzero limiting boundary value simply remains in the formula for the sum.

If the boundary sequence grows without bound, the partial sums in the corresponding identity cannot approach a finite value. If the boundary sequence oscillates without approaching a limit, the partial sums likewise fail to approach a finite value. In either case, the series diverges. The precise behavior can be different—partial sums may grow or oscillate—but convergence requires a finite limit.

This test is specific to a series whose terms have been expressed as differences of consecutive values. It does not say that every series can be decided by inspecting a boundary sequence; first, the telescoping representation must be established. When it is available, however, the finite identity makes the decision exact.

A Common Boundary-Term Mistake

A frequent error is to cancel the middle terms correctly and then discard the last boundary term as though it must vanish. It does not have to vanish. It must be evaluated through its limit.

For a concrete example, set $u_n=5+1/n$ and $a_n=u_n-u_{n+1}$. Then

a_n = \left(5+\frac1n\right) - \left(5+\frac{1}{n+1}\right) = \frac1n-\frac{1}{n+1} = \frac{1}{n(n+1)}.

An incorrect calculation might cancel the middle terms and claim that the sum is just $u_1=6$, by dropping the far boundary term. The correct finite sum and limit are

S_N=u_1-u_{N+1} = 6-\left(5+\frac{1}{N+1}\right) = 1-\frac{1}{N+1} \longrightarrow 1.

The far boundary approaches $5$, not $0$, so it must be subtracted. This agrees with the direct decomposition: the constant $5$ cancels within each difference, leaving the series from the first example, whose sum is $1$.

A Reliable Decision Process

When a series appears to telescope, keep the finite sum visible until the limit has been taken. The following sequence of steps prevents cancellation from hiding an important boundary term.

Identify the consecutive-value difference.
Write each term as $u_n-u_{n+1}$ or $u_{n+1}-u_n$, and verify the equality algebraically.

Form a finite partial sum.
Write the first and last terms explicitly, then cancel only terms that occur with opposite signs.

Examine the remaining boundary sequence.
Determine whether it approaches a finite limit, grows without bound, or fails to settle to a limit.

State convergence and the sum only after taking the limit.
Use the boundary limit in the finite identity; do not assume it is zero.

Why This Test Is Useful

Telescoping sums turn an apparently long collection of terms into a comparison between two boundary values. The important work is therefore not counting how many terms cancel, but identifying the sequence left at the moving end and determining its limit. This perspective also explains why the same finite cancellation pattern can produce a convergent series, an unbounded sequence of partial sums, or oscillating partial sums.

The finite identity is the source of every conclusion here. It gives the partial sums exactly, and the definition of series convergence asks whether those partial sums approach a finite limit. Once the two are connected, convergence and the value of the sum follow from the boundary behavior.

Check Your Understanding

Use a finite partial-sum identity before deciding each question.

If $a_n=u_n-u_{n+1}$ and $u_n\to 4$, what is the sum in terms of $u_1$?
For $a_n=u_{n+1}-u_n$, write the $N$th partial sum in terms of $u_1$ and $u_{N+1}$.
If $a_n=1/n-1/(n+1)$, identify the boundary sequence and determine the series sum.
Can a telescoping series converge when its boundary sequence approaches a nonzero number? Explain using the partial-sum identity.
If the boundary sequence oscillates between two distinct values, can the corresponding partial sums converge to a finite limit? Justify your answer.