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 $\forall n\in\mathbb N^*$:


*

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

*$\{a_n\}$ 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.
 A: Telescope! Note that $$ \frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}}&=
\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}\pars{n + 3}}
={\Psi\pars{3} - \Psi\pars{1} \over 3 - 1}\tag{1}
\end{align}
where $\Psi\pars{z}$ is the $\it digamma$ function. By using the identity
$\Psi\pars{z} = 1/\pars{z - 1} + \Psi\pars{z - 1}$ we'll get:
$$
\Psi\pars{3} = \half + \Psi\pars{2} = \half + 1 + \Psi\pars{1} = {3 \over 2} + \Psi\pars{1}
$$
We replace this result in $\pars{1}$:
$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}} = {3 \over 4}}
$$
