Summation Notation — Question 6

Question 6 - Solution

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

Notice that consecutive terms cancel: $$-\frac{1}{2}+\frac{1}{2}=0,\quad -\frac{1}{3}+\frac{1}{3}=0,\quad \ldots$$

After all cancellations, only the first and last terms remain: $$1-\frac{1}{n+1}.$$

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

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