Telescoping Series

The central idea

A telescoping series has partial sums in which most terms cancel after expansion, often following a decomposition $a_n=b_n-b_{n+1}$. Convergence and sum are determined by writing the finite partial sum $S_N$ first and then taking $N\to\infty$.

Definitions, hypotheses, and notation

When the shift exceeds one, several initial and final terms survive. For $b_n-b_{n+k}$, the first $k$ positive terms remain and are balanced by $k$ tail terms. Writing at least enough expanded terms to show the shift prevents accidental over-cancellation.

A telescoping series may still diverge if the surviving boundary expression lacks a finite limit. Conversely, not all convergent series telescope. The technique is an exact summation structure, so use it when cancellation is visible rather than as a generic convergence test.

Conceptual meaning

Cancellation is an exact finite phenomenon before it is a limiting one. Interior terms appear once positively and once negatively, leaving a few boundary terms. The surviving tail boundary controls the infinite sum.

A dependable method and decision rule

1. Use algebra or partial fractions to expose a difference pattern.
2. Write several terms with indices visible.
3. Form the finite partial sum through $N$.
4. Cancel only terms actually present and identify survivors.
5. Take the limit of the resulting $S_N$ formula.

Fully worked example

Graphical or geometric meaning

Think of adjacent transfers: each interior amount received from one term is removed by the next. Only the initial deposit and final unpaid tail remain. As the tail tends to zero, the initial amount becomes the sum.

Common mistakes and why they fail

Verification and reasonableness checks

• Compute $S_1$ and $S_2$ from both the original and closed formula.
• List surviving first and last terms explicitly.
• Confirm the remainder term has the correct limit.

Cancellation must be shown in partial sums

Rewrite a term as $b_n-b_{n+1}$, then expand the finite partial sum $S_N$. Middle terms cancel, leaving boundary terms whose limit determines both convergence and sum. Writing only an infinite cancellation can hide an index error. Partial fractions often produces the needed difference, but shifted subscripts must align exactly. If the surviving $N$-dependent term has no finite limit, the series diverges despite extensive cancellation. Finitely many uncanceled early terms affect the numerical sum even though they do not affect convergence. Verify the decomposition by recombining it and list several terms before taking a limit; both steps expose common sign and indexing mistakes.

For a tail beginning at an index other than one, expand from the actual lower bound. The surviving first boundary term changes with that bound, so copying the sum of a similar-looking series can give the right convergence verdict but the wrong numerical total.

Practice

1. Sum $\sum_{n=1}^\infty[1/n-1/(n+2)]$.
2. Find $S_N$ for $\sum_{n=1}^N1/[n(n+1)]$.
3. Does telescoping itself remove the need for a limit?

Connections and next steps

• Partial Fractions
• Infinite Series
• Geometric Series