MHB 10.02.10 Find the sum of the series

[karush]

$\textsf{Find the sum of the series}\\$
\begin{align*}\displaystyle
S_{n}&=\sum_{n=1}^{\infty}
\frac{4}{(4n-1)(4n+3)}=\color{red}{\frac{1}{3}} \\
\end{align*}
$\textsf{expand rational expression } $
\begin{align*}\displaystyle
\frac{4}{(4n-1)(4n+3)}
&=\frac{A}{(4n-1)}+\frac{B}{(4n+3)}\\
4&=A(4n+3)+B(4n-1)\\
\end{align*}
$\textit{if $\displaystyle n=-\frac{3}{4}$ then: }$
\begin{align*}\displaystyle
4&=-2B \therefore B=-1\\
\end{align*}
$\textit{if $\displaystyle n=\frac{3}{4}$ then: }$
\begin{align*}\displaystyle
4&=4A \therefore A=1\\
\end{align*}
$\textit{then}$
\begin{align*}\displaystyle
S_{n}&=\sum_{n=1}^{\infty}
\left[\frac{1}{(4n-1)}- \frac{1}{(4n+3)}\right]\\
\end{align*}
$\textit{partial sum $S_k$}$
\begin{align*}\displaystyle
&=\left[ \frac{1}{3}-\frac{1}{7} \right]
+\left[ \frac{1}{7}-\frac{1}{11} \right]
+\left[ \frac{1}{11}-\frac{1}{15} \right]
+ \cdots +
\end{align*}
$\textit{got lost here}$

 

[mathmari]

It holds that $$\frac{4}{(4n-1)(4n+3)} =\frac{1}{(4n-1)}-\frac{1}{(4n+3)}$$ We have the following partial sum:
$$S_{N}=\sum_{n=1}^N = \left[\frac{1}{4n-1}- \frac{1}{4n+3}\right]= \left (\frac{1}{3}- \frac{1}{7}\right )+\left (\frac{1}{7}- \frac{1}{11}\right )+\left (\frac{1}{11}-\frac{1}{15}\right )+\ldots +\left (\frac{1}{4N-1}- \frac{1}{4N+3}\right )$$

Every term will be canceled out except of the first and the last one. So, we get $$ S_N=\frac{1}{3}- \frac{1}{4N+3}$$

Therefore, $$\sum_{n=1}^{\infty}=\lim_{N\rightarrow \infty}S_N=\lim_{N\rightarrow \infty}\left (\frac{1}{3}- \frac{1}{4N+3}\right )=\frac{1}{3}$$

 

[karush]

thanks

guess the last few steps alluded me