<s>[INST] A periodic summation question 
The periodic series can be solved by doing 

I understand the solution but it is little lengthy process. Can't we apply any law to get the solution direct?
 [/INST] Splitting the sum and performing an index shift gets you there pretty quick.
$$
\begin{align*}
\sum_{n=1}^{20} a_n
&= \sum_{n=1}^{20} \left(\frac{1}{n}-\frac{1}{n+2}\right) 
= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=1}^{20} \frac{1}{n+2} \\
&= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=3}^{22} \frac{1}{n} 
= \sum_{n=1}^{2} \frac{1}{n} - \sum_{n=21}^{22} \frac{1}{n} \\
&= \left(\frac 1 1+\frac 1 2\right) - \left(\frac 1 {21} + \frac 1 {22}\right)
\end{align*}
$$
This is a typical case of a telescoping sum as mentioned by Sharkos.
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