Calculate the sum of the series: $\sum\limits_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )$ I am stuck at a part when it comes to evaluating the sum of the said series... Here is my work so far (and I am not sure if the notation and simplification is correct either):
Simplifying using partial sums:
$$\sum_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )=\sum_{n=1}^{\infty}\left ( 3\left ( \frac{1}{n}-\frac{1}{n+1} \right )-\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right ) \right )$$
Now I take the limit of the Nth partial sum of the series, right?
$$\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( 3\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) \right )$$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) $$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{N}-3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{N+1} -\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{2N+3}+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{2N+5} $$
I assume I doing something wrong here, because each term diverges. This is what was written in the textbook:
$$\sum_{n=1}^{\infty}\left ( \frac{1}{n}-\frac{1}{n+1} \right )=1-\lim_{N\rightarrow \infty}\frac{1}{N}=1$$
$$\sum_{n=1}^{\infty}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )=\frac{1}{5}-\lim_{N\rightarrow \infty}\frac{1}{2N+3}=\frac{1}{5}$$

The problem is...I am not sure how they got this! They are missing a lot of steps for me to understand, hence the messiness above.

 A: Note that both series $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n}~~and ~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15}$$
converges then there is no risk to separate the sum.  
However using  telescoping sum for the second sum as follows let $u_n=\frac{1}{2n+3}$ then $u_{n+1}=\frac{1}{2n+5}$ hence
$$ ~~~~\sum_{n=1}^{\infty}\frac{2}{4n^{2}+16n+15} =\lim_{k\to\infty }\sum_{n=1}^{k}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )\\\color{red}{:=\lim_{k\to\infty }\sum_{n=1}^{k}\left ( u_n-u_{n+1} \right )=\lim_{k\to\infty }(u_1-u_{k+1})}\\=\frac{1}{5}-\lim_{k\rightarrow \infty}\frac{1}{2k+5}=\frac{1}{5}$$
whereas the first is obvious since 
$$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n} = 3\lim_{k\to\infty }\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) =3\lim_{k\to\infty }(1-\frac{1}{k+1}) = 3~~$$
and  
A: Hint:
The given expression can be written as 
$$\sum_{n=1}^\infty 
\frac {3}{n(n+1)} - \frac {2}{(2n+3)(2n+5)}$$
On partial decomposition it becomes 
$$\sum_{n=1}^\infty  \left[3\left(\frac {1}{n}-\frac {1}{n+1}\right)  - \left(\frac {1}{2n+3}-\frac{1}{2n+5}\right) \right]$$
Can you see the series telescoping. By the way the answer I guess might be $\frac {14}{3}$
