Integral Test for $\sum_{n=1}^{\infty}n\left(1+n^{2}\right)^{p}$ I am trying to work through the following textbook exercise:
Find the values of p for which the series converges:
$$
\sum_{n=1}^{\infty}n\left(1+n^{2}\right)^{p}
$$
I have found a number of videos and explanations on how to use the integral test to solve this - which is convergent for $p\lt -1$.
However, they all gloss over meeting the three criteria to use the integral test.
Positive and continuous on the interval $\left[ 1,\infty  \right)$ is clear.
But I am stuck on showing it is eventually decreasing on the interval.
When I take the derivative of $f(x)$, I get:
$$
f'(x)=2px^{2}\left(x^{2}+1\right)^{p-1}+\left(x^{2}+1\right)^{p} 
$$
What am I missing as I can't see how it is decreasing? And therefore, how the integral test can be used.
 A: It is unfortunate, but this part of the test is often swept under the rug. I have a similar complaint about most uses of the alternating series test. Good on you for demanding some proof!
We have,
\begin{align*}
f'(x) &= 2px^2\left(x^2+1\right)^{p-1}+\left(x^2+1\right)^p \\
&= 2p\left(x^2 + 1\right)\left(x^2+1\right)^{p-1}+\left(x^2+1\right)^p-2p\left(x^2 + 1\right)^{p-1} \\
&= (2p+1)\left(x^2+1\right)^p-2p\left(x^2 + 1\right)^{p-1}.
\end{align*}
We have $f'(x) < 0$ if and only if
$$(2p+1)(x^2 + 1) < 2p \iff (2p + 1)x^2 < -1.$$
If we assume that $p < -\frac{1}{2}$ (which would, of course, be true if $p < -1$), then this is equivalent to,
$$x > \frac{1}{\sqrt{|2p + 1|}},$$
which is what we need. So, for sufficiently large $x$, provided $p < -\frac{1}{2}$, the function is decreasing. If $p \ge -\frac{1}{2}$, then looking back at the previous working, it's clear that the function is actually increasing everywhere. This shows both that the integral and the series will not converge (as both increase from a positive number), but also that the test technically doesn't apply in that case.
Hope that helps!
A: It is easy to prove that $f'(x)<0 $ for $x> N$ if  $\,p< -1$. So the integral converges and hence the series converges for $\,p< -1$. For $p=-1$ the derivative is negative for $x> N$ so the test can be applied but the integral and hence the series diverges. For $-1<p<-\frac{1}{2}$ the derivative is negative for $x> N\,\,$ so the test can be applied and give that the integral and hence the series diverges. For $p\geq -\frac{1}{2}$ it is clear that $n(1+n^{2})^{p}\geq n(1+n^{2})^{-1/2}$ and $\frac{n}{\sqrt{1+n^2}}> \frac{1}{\sqrt{1+n^{2}}}>\frac{1}{n+1}$ and the last series diverges. Thus the series converges for $p<-1$ and diverges for $p\geq 1$.!!
A: Since
$$
I_p=\int_1^{\infty}x(1+x^2)^pdx=\frac12\int_2^{\infty}u^pdu=\begin{cases}
-\frac{2^p}{p+1} \text{ for } p <-1\cr
\infty \text{ for } p \ge -1
\end{cases}. 
$$
the series converges for $p<-1$ and diverges otherwise.
