The right way to cancel out the terms in the following telescoping series So how do I cancel and simplify the terms in the following telescopic series.
Been at it for hours, cant seem to figure it out.
$\sum\limits_{k = 1}^n \frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)} $
Any help would be deeply appreciated.
P.S: I need to show that its equal to the following,
$\frac{1}{12} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}$
But I cant seem to figure out the right cancellation method.
 A: Here is another variant

\begin{align*}
\sum_{k=1}^n&\left(\frac{1}{2(k+1)}-\frac{1}{k+2}+\frac{1}{2(k+3)}\right)\\
&=\frac{1}{2}\sum_{k=1}^n\frac{1}{k+1}-\sum_{k=1}^n\frac{1}{k+2}+\frac{1}{2}\sum_{k=1}^n\frac{1}{k+3}\tag{1}\\
&=\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k+1}-\sum_{k=2}^{n+1}\frac{1}{k+1}+\frac{1}{2}\sum_{k=3}^{n+2}\frac{1}{k+1}\tag{2}\\
&=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{3}+\frac{1}{n+2}\right)+\frac{1}{2}\left(\frac{1}{n+2}+\frac{1}{n+3}\right)\tag{3}\\
&=\frac{1}{12}-\frac{1}{2(n+2)}+\frac{1}{2(n+3)}\tag{4}
\end{align*}

Comment:


*

*In (1) we split the sum

*In (2) we shift the index by $1$ resp. $2$ 

*In (3) we observe that the sums with index range $k=3$ up to $k=n$ cancel away

*In (4) we collect terms
A: \begin{eqnarray*}\sum_{k=1}^n \frac1{2(k+1)}-\frac1{k+2}+\frac1{2(k+3)}&=&\\
&=&\frac14+\frac16+\frac1{2n+4}-\frac1{2n+6}+\sum_{k=3}^{n}\frac1{k+1}-\sum_{k=0}^{n-1}\frac1{k+1}\\
&=&\frac14+\frac16+\frac1{2n+4}-\frac1{2n+6}+\frac1{n+1}-\frac11-\frac12-\frac13
\end{eqnarray*}
A: We have:
$$
a_n=\sum\limits_{k = 1}^n \left(\frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)}\right)= a_{n-1}+ \left(\frac{1}{2(n+1)} -\frac{1}{n+2}+\frac{1}{2(n+3)}\right)=a_{n-1}+\frac{1}{(n+1)(n+2)(n+3)}
$$
Does this solve your problem?
A: For a telescoping sum you would like to have the trivial form $a_k=b_k-b_{k+1}$. By comparing the largest and smallest terms of $a_k$ one finds at second glance that
$$
b_k=\frac1{2(k+1)}-\frac1{2(k+2)}=\frac1{2(k+1)(k+2)}
$$
fits the bill. Thus
$$
\sum_{k=1}^n a_k=b_1-b_{n+1}=\frac1{2·2·3}-\frac1{2(n+2)(n+3)}
$$
