Find all values of $p$ for which the series $\sum \frac{n}{(n^2-1)^p}$ converges 
Use the integral test to find all values of p for which the series converges
$$\sum a_n =\sum \frac{n}{(n^2-1)^p}$$

The interval is chosen such that $a_n$ is defined $\implies n \geq 2$.
(a) Condition for the integral test:
For the integral test to work, the function has to be decreasing and positive. As $a_n$ is a function of $n$, let's consider the function $$f(x) = \frac{x}{(x^2-1)^p}$$.
$$f'(x)= (x^2-1)^{-p-1}(x^2(-2p+1)-1)$$
two critical points exist in the reals $x=1, x=-1$.Testing  $f'$ in the interval of interest $[2, \infty)$,
As we want $f$ to be decreasing, so $P$ is such that $$p > \frac{x^2-1}{2x^2}$$
$\forall x \in [2, \infty)$, $x^2 -1 > 0 \implies f(x)>0$, $\forall p$.
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(b) the integral
$$\int_{2}^{\infty} x(x^2-1)^{-p}= \frac{1}{2} \lim\limits_{t \rightarrow \infty} \left[\frac{1}{1-p} (x^2-1)^{1-p} \right]_2^t = \frac{1}{2} \lim\limits_{t \rightarrow \infty} \left( \frac{(t^2-1)^{1-p}}{1-p} - \frac{3^{1-p}}{1-p} \right)$$
We can observe that the integral is indeterminate when $p=1$
$$ \int f <\infty \implies   \lim\limits_{t \rightarrow \infty}  \frac{(t^2-1)^{1-p}}{1-p} <\infty \implies \lim\limits_{t \rightarrow \infty}  (t^2-1)^{1-p} <\infty \implies 1-p \leq 0 \implies p\geq 1 $$
To conclude $\sum a_n < \infty$ iff ${p \geq 1}$ and $p>\frac{x^2-1}{2x^2}$
.
I would like to know if my argumentation is correct, appropriate for the integration test?
Much appreciated for your input/help
 A: The easiest approach to determining convergence of this series would be to invoke the Limit Comparison test between this series and $\sum \frac{n}{n^{2p}}=\sum \frac{1}{n^{2p-1}}$, since for the latter we know precisely when it converges of diverges. But if you have to apply the Integral Test directly showing all work, then your solution is mostly correct… except for some errors and inaccuracies in it.

For the integral test to work, the function has to be decreasing

Absolutely correct! That's why you should immediately conclude that $p>0$, since otherwise the terms are not decreasing. So from now on, we're assuming that $p>0$.

Two critical points exist in the reals $x=1$, $x=−1$.

That depends. From the second factor of $f'(x)$, we can also have $x^2=\frac{1}{1-2p}$. So if $p<1/2$, then we have two more critical points $x=\pm\frac{1}{\sqrt{1-2p}}$. But fortunately that doesn't matter: $x=1$ is still the right-most critical point, and we can check that regardless of $p$ (however, remember that our $p>0$) the derivative is negative on $[2,+\infty)$.

We can observe that the integral is indeterminate when $p=1$

No, that's not what's going on here. Your whole calculation simply doesn't work (doesn't apply) when $p=1$. It's a completely different integral in that case. So instead you should keep the calculation that you did (and you did it correctly!), but with a comment that this is for the case when $p>0$ and $p\neq1$. Then you will determine, as you did, that the series diverges for $p<1$ and converges for $p>1$. But the case $p=1$ has to be redone separately, as the antiderivative in this case is completely different — a logarithm, not a power function.
A: HINT.
For $1<n\in \mathbb N$ and $p>1/2$ we have $0<1/(n^2-1)<1/(n^2/2)^p.$
For $1<n\in \mathbb N$ and $p\leq 1/2$ we have $1/(n^2-1)^p>1/(n^2)^p.$
A: $\int \frac{n}{\left(n^2-1\right)^p} \, dn=\frac{1}{(2-2 p) \left(n^2-1\right)^{p-1}}$
so the condition for the convergence is $p>1$
If $p=1$ the integral diverges because $\frac{n}{n^2-1}$ is asymptotic to $\dfrac{1}{n}$ as $n\to +\infty$
