Summation Notation — Question 7

Question 7

Prove that for any positive integer $n$, $$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{n}{n+1}.$$

Question 7 - Solution

Begin by rewriting the general term using partial fractions.

For each $k\ge 1$, $$\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}.$$

Substitute this into the sum: $$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right).$$

Write out the first few terms: $$\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) +\cdots+ \left(\frac{1}{n}-\frac{1}{n+1}\right).$$

All intermediate terms cancel, leaving $$1-\frac{1}{n+1}.$$

Simplify: $$1-\frac{1}{n+1} = \frac{n}{n+1}.$$

$$\boxed{\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}}$$