Proving a series converges and finding its sum I have the series:
$$\sum_{n=1}^{\infty}\frac {1}{25n^2 + 5n - 6}$$ and I am supposed to prove that the series converges and then find its sum. I have to do multiple problems like this. Can I get an example of how to solve problems like these? You don't have to solve this problem in particular if you don't want. I feel an example problem would help immensely though.
EDIT: I am so sorry. I entered the wrong series. I was looking the problem where it was indeed just prove that the series converges. The series has been updated to one of the question where both proving that it converges and finding the sum is the requirement.
This was what I had initially and since a clue was given I thought I might as well think about it too. 
$$\sum_{n=1}^{\infty}\frac {n+2}{n^3 + 3n^2 + 1}$$ 
However, I can seem to rewrite the series so I can compare it to $\frac {1}{n^2}$. The denominator can't be factored further can it? 
EDIT 2: Attempt at what I originally had.
$$\begin{align}&\sum_{n=1}^{\infty}\frac {n+2}{n^3 + n^2 +1}  \\
&\le\sum_{n=1}^{\infty}\frac {n+2}{n^3}  \\
&=\sum_{n=1}^{\infty}\frac {n}{n^3}+\sum_{n=1}^{\infty}\frac {2}{n^3}
\end{align}$$ 
Since these two summations converge then the original must converge as well by the comparison test. 
How is this proof of the original equation, does it work? I haven't really written a proof using a comparison test before so I'm not sure if it needs to be more detailed or not.
 A: I suspect "find its sum" is meant in  the sense of numerical approximation, not a closed-form formula. The best closed form formula in this case seems to be
$$ \sum_r \frac{r (3 + r)}{3} \Psi(1-r) $$
where the sum is over the roots of the polynomial $x^3 + 3 x^2 + 1$, and $\Psi$ is the digamma function.
EDIT: That was for $$\sum_{n=1}^\infty \frac{n+2}{n^3+3 n^2+1}$$
The new series $$\sum_{n=1}^\infty \frac{1}{25 n^2 + 5 n - 6}$$
is a telescoping series, since 
$$ \frac{1}{25 n^2 + 5 n - 6} = \frac{1}{5 (5n-2)} - \frac{1}{5(5(n+1)-2)}$$
As for the comparisons to the series $\sum_n \frac{1}{n^2}$, the simplest way of doing that is with the Limit comparison test.
A: Actually it is pretty easy to find the sum of the given series, once noticed that:
$$ \sum_{n=1}^{N}\frac{1}{25n^2+5n-6} = \frac{1}{5}\sum_{n=1}^{N}\left(\frac{1}{5n-2}-\frac{1}{5n+3}\right) = \frac{1}{5}\left(\frac{1}{3}-\frac{1}{5N+3}\right) $$
is a telescopic sum. The given series simply equals $\color{red}{\large\frac{1}{15}}$.
A: Notice: I was typing while other reply is posted. Anyway, I decided to leave my reply for the details.
The sum can be calculated by telescoping. The answer is $\frac{1}{15}.$
Here are the details:
The partial sum is:
$$S_k=\sum_{n=1}^k \frac{1}{25n^2+5n-6}=\sum_{n=1}^k \frac{1}{(5n-2)(5n+3)}=\sum_{n=1}^k \frac{1}{5}\left(\frac{1}{5n-2}-\frac{1}{5n+3}\right)=\frac{1}{5}\left[\left(\frac{1}{3}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{18}\right)+...+\left(\frac{1}{5k-2}-\frac{1}{5k+3}\right)\right]=$$
$$\frac{1}{5}\left(\frac{1}{3}-\frac{1}{5k+3}\right).$$
The sum of the series is the limit of its partial sum:
$$\sum_{n=1}^\infty \frac{1}{25n^2+5n-6}=\lim_{k\to \infty} S_k=\lim_{k\to \infty} \frac{1}{5}\left(\frac{1}{3}-\frac{1}{5k+3}\right)=\frac{1}{15}.$$
