Series $\sum_\limits{n=0}^\infty \frac{n+1}{n^3-7}$ 
I would like to prove the series $\sum_\limits{n=0}^\infty\frac{n+1}{n^3-7}$ is convergent. 

I have tried the ratio test but it is inconclusive, what is the way to go here?
Thanks
 A: Observe that for large $n$ you have that $n+1\approx n$ and $n^3-7\approx n^3$ and therefore the ratio $\displaystyle \frac{n+1}{n^3-7}$ behaves like $\displaystyle \frac{n}{n^3} =\frac{1}{n^2}$, whose series $\displaystyle \sum\frac{1}{n^2}$ converges (it is well known that the $p-$series 
$\displaystyle \sum\frac{1}{n^p}$ converges if and only if $p>1$), so your series also converges. 
If you'd like to use a formalistic path, observe that for large enough $n$ you have $\displaystyle \frac{n+1}{n^3-7}<\frac{2}{n^2}$ (just turn it to a non-ratio expression to convice yourself) and use the comparison test. 
Yet another way is to use the Limit Comparison Test, by noting that for $a_n=\displaystyle \frac{n+1}{n^3-7}$ and $\displaystyle b_n=\frac{1}{n^2}$ we have $\dfrac{a_n}{b_n}\rightarrow 1$ and the series $\displaystyle \sum\frac{1}{n^2}$ converges.
A: You can compare the series to $\sum \limits_{n=1}^{\infty} \frac{1}{n^2}$, which we know converges. 
A: Hint: find a constant $C > 0$ so that $a_n < \dfrac{C}{n^2} $ and use comparison test.
A: This is a comment,
but easier to enter as an answer.
The ratio test does not work because,
for any sum of the form
$\sum_{n=1}^{\infty} \frac1{n^a}$
with $a > 1$
(which this is),
the ratio is
$\frac{\frac1{(n+1)^a}}{\frac1{n^a}}
=\frac{n^a}{(n+1)^a}
=\frac1{(1+1/n)^a}
\to 1
$
as $n \to \infty$.
Similarly,
the root test fails because
$\left(\frac1{n^a}\right)^{1/n}
=\frac1{n^{a/n}}
=\frac1{(n^{1/n})^a}
\to 1
$
because
$n^{1/n} \to 1$.
Now you try the
integral test and Cauchy's condensation test
and see what happens.
