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?

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.
