Sequences and Series – Infinite Series Sum of 1/(n(n+1))

sequences-and-series

I am confused on the following series:

$$\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)} = 1$$

My calculator reveals that the answer found when evaluating this series is 1. However, I am not sure how it arrives at this conclusion. I understand that partial fractions will be used to create the following equation. I just don't understand how to proceed with the problem.

$$\sum\limits_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1$$

Best Answer

Write out a few terms of the series. You should see a pattern! But first consider the finite series:

$$\sum\limits_{n=1}^{m}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{m-1} - \frac{1}{m} + \frac{1}{m} - \frac{1}{m+1}.$$ This sum is telescoping, since it collapses like a telescope.

Everything is left except for the first and last term. Now what's the limit as $m\to \infty$?