# Why $\lim_{n\to \infty}{\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\dots+\frac{1}{n\cdot(n+2)}}=\frac{3}{4}$?

Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{1}{1\cdot3}+\cfrac{1}{2\cdot4}+\dots+\cfrac{1}{n\cdot(n+2)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.

I write: $$\lim_{n\to \infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n\cdot(n+2)}}=\sum_{n=1}^{\infty}{\frac{1}{n^2+2n}}\approx\sum_{n=1}^{\infty}{\cfrac{1}{n^2}}$$

I put some values of $n$ for finding a pattern: $$\begin{array}{c|lcr} n & \text{1}&\text{2}&\text{3}&\text{4}\\ \hline \sum &\cfrac{1}{3}&\cfrac{11}{24}&\cfrac{21}{40}&\cfrac{17}{30} \end{array}$$

... but no hope. I know that the limit/series converges, because $\forall n\in\mathbb N^*$:

1. $\{a_n\}$ is increasing by the test of monothony : $a_{n+1}-a_{n}=\cfrac{1}{(n+1)(n+3)}>0$

2. $\{a_n\}$ is bounded : $0\le a_1=\frac{1}{3}\le a_2=\frac{11}{24}\le \dots \le a_n \le1$

Wolfram says that the summation can be written as follows: $$\cfrac{3}{4}-\cfrac{2n+3}{2(n+1)(n+2)}$$ How did it end up at this formula? Thank you.

• Yes, yes! My mistake, thank you! – Daniel C Jan 18 '14 at 9:34
• Pronic numbers for more... – Fred Kline Jan 18 '14 at 9:42

Telescope! Note that $$\frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$