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

Question

Just like i know there is a formula that gives the following: $$\sum_{i=1}^k i =\frac{k(k+1)}{2}$$ I want to know if there's a formula for this sum: $$\sum_{i=1}^k \frac{i}{(i+1)}$$

I haven't been able to find it on the web.

Answer 1

Because $\lim_{n\rightarrow\infty}\frac{n}{n+1}=1$, the sum

$$\sum_{n=1}^\infty\frac{n}{n+1}$$

cannot converge.

Note that

$$\begin{align*} \frac{n}{n+1} &=\frac{n+1-1}{n+1} \\ &=\frac{n+1}{n+1}-\frac{1}{n+1} \\ &=1-\frac{1}{n+1}. \end{align*}$$

So

\begin{align*} \sum_{n=1}^k\frac{n}{n+1} &=k-\sum_{n=1}^k\frac{1}{n+1} \\ &=k-H_{k+1}+1 \end{align*}

where $H_m$ is the mth harmonic number.