Calculating limit of series. Given a series $\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}$, how do I find the limit? I understand I need to find the sequence of partial sums, which goes something like this $s_n=\{\frac{5}{24}, \frac{7}{30}, \frac{27}{112}....\}$which will probably converge at 0.25, but I am having trouble finding a description of the sequence on which I could evaluate the limit.
 A: As a hint:$$\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{4n+1}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{(2n-1)+(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{(2n-1)}{(2n-1)(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\
\sum_{n=1}^{\infty}\frac{1}{(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+0)(2n+1)}=\\$$ and now you have two telscopic series.can you take over now ?
A: The partial fraction decomposition of the summand is
$$\frac1{2(2n-1)}-\frac1{2(2n+1)}-\frac1{4n}+\frac1{4(n+1)}$$
The first and second parts, and the third and fourth, telescope. In the infinite sum all parts except the $n=1$ instances of the first and third parts will cancel, so the infinite sum is
$$\frac1{2(2×1-1)}-\frac1{4×1}=\frac14$$
A: Let $\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=f(n-1)-f(n)$  where $f(r)=\dfrac{ar+b}{(2r+1)(2r+2)}$
so that $$\sum_{n=1}^m\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=\sum_{n=1}^m(f(n-1)-f(n))=f(0)-f(m)$$
We need
$f(n-1)-f(n)=\dfrac{(2n+1)(2n+2)(an-a+b)-(an+b)(2n-1)2n}{(2n-1)2n(2n+1)(2n+2)}$
$\implies4n+1=n^2(4a)+2n(?)+2b-2a$
Comparing the coefficients of $n^2,4a=0\iff a=?$
Comparing the constants, $2b-2a=1\implies b=\dfrac12$
These values of  $a,b$ satisfy the coefficients of $n$
Now set $m\to\infty$ and $\lim_{m\to\infty}f(m)=?$
