Technique

Telescoping a sum by writing each term as a difference

Algebra • Sequence and Series • Summation

Statement

If every term can be written as

f (n) - f (n + 1)

, then

\sum_{n = a}^{b} (f (n) - f (n + 1)) = f (a) - f (b + 1)

, because each interior value is added once and subtracted once.

How to use it

1Split a rational term by partial fractions first: $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ .
2Write out the first two and the last two terms to see exactly which pieces survive.
3If each term is a constant plus a difference, count the constants separately: they contribute the constant times the number of terms.

Evaluate

\sum_{n = 1}^{99} \frac{1}{n ( n + 1 )}

Step-by-step

1
$\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$
Why: Split the term so that consecutive terms cancel.
2
$\sum_{n = 1}^{99} (\frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{1} - \frac{1}{100}$
Why: Every interior fraction appears once with each sign.
3
$= \frac{99}{100}$

Answer

\frac{99}{100}

Watch out

The surviving tail is

f (b + 1)

, not

f (b)

: stopping the cancellation one index early is the commonest error, and it is exactly what a well-built option list punishes.

Also known as

telescoping series, method of differences, cancelling sum, partial fractions sum