<s>[INST] 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.
 [/INST] 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
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