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 $$p &gt; \frac{x^2-1}{2x^2}$$
$\forall x \in [2, \infty)$, $x^2 -1 &gt; 0 \implies f(x)&gt;0$, $\forall p$.
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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 $p=1$
$$ \int f &lt;\infty \implies   \lim\limits_{t \rightarrow \infty}  \frac{(t^2-1)^{1-p}}{1-p} &lt;\infty \implies \lim\limits_{t \rightarrow \infty}  (t^2-1)^{1-p} &lt;\infty \implies 1-p \leq 0 \implies p\geq 1 $$
To conclude $\sum a_n &lt; \infty$ iff ${p \geq 1}$ and $p&gt;\frac{x^2-1}{2x^2}$
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I would like to know if my argumentation is correct, appropriate for the integration test?
Much appreciated for your input/help
HINT.

For $1&lt;n\in \mathbb N$ and $p&gt;1/2$ we have $0&lt;1/(n^2-1)&lt;1/(n^2/2)^p.$

For $1&lt;n\in \mathbb N$ and $p\leq 1/2$ we have $1/(n^2-1)^p&gt;1/(n^2)^p.$
$\int \frac{n}{\left(n^2-1\right)^p} \, dn=\frac{1}{(2-2 p) \left(n^2-1\right)^{p-1}}$

so the condition for the convergence is $p&gt;1$

If $p=1$ the integral diverges because $\frac{n}{n^2-1}$ is asymptotic to $\dfrac{1}{n}$ as $n\to +\infty$