AP® Calculus BC · Unit 10 · Topic 10.1

Partial Sums and Telescoping Series

A series' sum is defined as the limit of its partial sums. A telescoping series is one where terms cancel in a chain, leaving only a few survivors so that limit is easy to compute.

To add infinitely many numbers, you never actually add infinitely many numbers at once. You build partial sums: S1 = a1, S2 = a1 + a2, S3 = a1 + a2 + a3, and so on. The infinite series is defined to converge if this sequence of partial sums approaches a single number L as n goes to infinity. That limit L is the sum. If the partial sums grow without bound or bounce around forever, the series diverges and has no sum.

Telescoping series are the cleanest case because you can find a formula for Sn directly instead of guessing a pattern. The trick is rewriting each term as a difference of two pieces that overlap with neighboring terms, like 1/(n(n+1)) = 1/n - 1/(n+1), using partial fractions. When you list out the partial sum, every interior term appears once with a plus sign and once with a minus sign, so it cancels. What's left is just the first leftover piece and the last leftover piece. Taking the limit as n goes to infinity often sends the last piece to zero, leaving a clean finite answer.

This topic is the foundation for everything else in the unit: every convergence test you learn later is really a way of predicting the behavior of partial sums without being able to compute them exactly, the way telescoping lets you here. Geometric series, which come next, are the other major family where Sn has a closed form you can take a limit of directly.

Key terms

Partial sum
The sum of the first n terms of a series, written Sn.
Convergent series
A series whose sequence of partial sums approaches a finite limit.
Divergent series
A series whose partial sums do not approach any finite limit.
Telescoping series
A series whose terms can be rewritten as differences so that most terms cancel in the partial sum.
Partial fraction decomposition
Rewriting a fraction like 1/(n(n+1)) as a sum or difference of simpler fractions.

Watch it explained

  1. How it worksHow does splitting 1/(n(n+1)) into 1/n minus 1/(n+1) make the whole sum telescope?
  2. Worked exampleFind the sum from n = 1 to infinity of 1/(n(n + 1)).
  3. Worked exampleFind the sum from n = 1 to infinity of the difference 1/n minus 1/(n + 2).

Check yourself

Your turn1 of 2
Find the sum from n = 1 to infinity of 7/(n(n + 1)).

End of topic.