Telescoping Sums

Write each term as a_k - a_k+1 and watch the whole sum collapse.

The idea

Telescoping evaluates a long sum in closed form by writing each term as a difference. Suppose the $k$th term can be written as $a_{k} - a_{k+1}$ for some sequence $a_{k}$. Then consecutive terms cancel: the $-a_{k+1}$ ending one term meets the $+a_{k+1}$ beginning the next, and only the two ends survive:

$\sum_{k=1}^{n} (a_{k} - a_{k+1}) = a_{1} - a_{n+1}.$

A sum of $n$ terms collapses to a single subtraction.

The work lies in finding the sequence $a_{k}$. When each term is a fraction whose denominator factors into consecutive pieces, splitting it into partial fractions supplies the difference. Products telescope the same way, with ratios in place of differences: if each factor can be written as $a_{k+1}/a_{k}$, the product collapses to $a_{n+1}/a_{1}$.

Reach for telescoping when the terms have a factored denominator or are visibly a difference of two similar expressions, and especially when the problem asks for a closed form in $n$ rather than a single number, since telescoping delivers a formula.

Ways to work on it

• Walkthrough. Split each term into a difference of consecutive fractions and collapse a long sum.
• Practice. Split the general term and collapse telescoping sums of varying shape and length.
• Hardest. Evaluate a telescoping sum with a wider cancellation gap, or a telescoping product.

Not sure where to start? Take the ten-question placement test.