# 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 $$\frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\pars{1}$: $$\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}} = {3 \over 4}}$$