Summation Notation — Question 9

Question 9

Prove that for every positive integer $n$, $$\sum_{k=1}^n\frac{k}{k+1}=n-\sum_{k=1}^n\frac1{k+1}.$$

Question 9 - Solution

For every $k\geq1$,

$$\frac{k}{k+1}=1-\frac1{k+1}.$$

Summing all $n$ terms gives

$$\boxed{\sum_{k=1}^n\frac{k}{k+1}=n-\sum_{k=1}^n\frac1{k+1}.}$$

Equivalently, the result is $n+1-H_{n+1}$, where $H_{n+1}=\sum_{j=1}^{n+1}1/j$.

For $n=1$, both sides of the corrected identity equal $1/2$.