Find the limit of \(S_n=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{(2n+1)(2n+3)}\).

Solution

The general term of this sequence is:

$$ \frac{2}{(2k+1)(2k+3)}=\frac{1}{2k+1}-\frac{1}{2k+3}. $$

Therefore,

$$ \begin{aligned} S_n&=\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\cdots+\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\\ &=1-\frac{1}{2n+3}. \end{aligned} $$

When \(n\rightarrow\infty\), we have \(\frac{1}{2n+3}\rightarrow0\).

Hence,

$$ \boxed{\lim_{n\to\infty}S_n=1}. $$

Solution by: Thin Sokkean