Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following: 
$$
\frac 1 {(2k+1)(2k+5)} = \frac a {(2k+1)} + \frac b {(2k+5)} \\~\\
\frac 1 {(2k+1)(2k+5)} = \frac {a(2k+5)} {(2k+1)(2k+5)} + \frac {b(2k+1)} {(2k+1)(2k+5)}  \\~\\$$
Then we have: 
$$
k: 2a + 2b = 0 \\
k^0: 5a + b = 1 \\~\\
a = \frac 1 4 ~ \land b = -\frac 1 4
$$
So we have: 
$$
\frac 1 {(2k+1)(2k+5)} = \frac { \frac 1 4 } {(2k+1)} - \frac { \frac 1 4} {(2k+5)} = \\ \frac { 1 } {4(2k+1)} - \frac { 1 } {4(2k+5)} 
$$
Which can be defined as:
$$
\frac 1 4 (\frac { 1 } {2k+1} - \frac { 1 } {2k+5} )
$$

I'm not sure how to turn the result into a telescopic series (if the partial fraction decomposition is right in the first place). 

 A: Yes, this sum telescopes. We have: $$S =\sum_{k=2}^{\infty}\frac14\left(\frac1{2k+1}-\frac1{2k+5}\right)$$ giving us $$4S =\left(\frac15-\frac19\right) +\left(\frac17-\frac1{11}\right)+\left(\frac19-\frac1{13}\right)+ \ldots$$ $$\implies 4S=\frac15+\frac17$$ $$\boxed{\therefore S=\frac3{35}}$$
A: You are on the right track, this is a telescopic series. Thanks to your partial fraction decomposition (which is correct), we have that for $n\geq 2$,
$$
\begin{align}\sum_{k=2}^n \frac 1 {(2k+1)(2k+5)}&=
\frac 1 4 \sum_{k=2}^n\left(\frac { 1 } {2k+1} - \frac { 1 } {2k+5} \right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=2}^n\frac { 1 } {2k+5}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=4}^{n+2}\frac { 1 } {2(k-2)+5}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=4}^{n+2}\frac { 1 } {2k+1}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^3\frac { 1 } {2k+1}-\sum_{k=n+1}^{n+2}\frac { 1 } {2k+1}\right)\\
\end{align}$$
where in the second sum we shifted the index.
Can you take it from here?
A: Set $u_{k} =\frac { 1 } {2k+1}\implies  u_{k+1} =\frac { 1 } {2k+3}~~ãnd ~~~u_{k+2} =\frac { 1 } {2k+5} $  then by telescopic sum we have 
$$\begin{align}\sum_{k=2}^n \frac 1 {(2k+1)(2k+5)}&=
\frac 1 4 \sum_{k=2}^n\left(\frac { 1 } {2k+1} - \frac { 1 } {2k+5} \right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+2}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+1}+u_{k+1}-u_{k+2}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+1}+\sum_{k=2}^nu_{k+1}-u_{k+2}\right)\\
&=\frac 1 4 \left(u_2-u_{n+1}+u_{3}-u_{n+2}\right)\to \frac 1 4 \left(u_2+u_{3}\right)\\&= \frac{3}{35}\end{align}$$
A: To prove the convergence of the series we can use the limit comparison test. Indeed
$$
\frac{1}{(2k+1)(2k+5)}\sim\frac{1}{4k^2}.
$$
Since $\sum_{1}^\infty k^{-2}$ converges it follows that 
$$
\sum_{k=2}^\infty\frac1{(2k+1)(2k+5)}<\infty.
$$
