Summation Notation — Question 10

Question 10

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

Question 10 - Solution

Write both sums explicitly: $$\sum_{k=1}^{n} \frac{1}{k} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n},$$ $$\sum_{k=1}^{n} \frac{1}{k+1} = \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n+1}.$$

Subtract the second sum from the first: $$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) - \left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right).$$

All common terms cancel: $$\frac{1}{2}-\frac{1}{2}=0,\quad \frac{1}{3}-\frac{1}{3}=0,\quad \ldots$$

Only the first term of the first sum and the last term of the second sum remain: $$1-\frac{1}{n+1}.$$

Thus, $$\sum_{k=1}^{n} \frac{1}{k}-\sum_{k=1}^{n} \frac{1}{k+1} = 1-\frac{1}{n+1}.$$

$$\boxed{1-\frac{1}{n+1}}$$