Finding $p$ such that $\sum\limits_{n=1}^\infty n(1+n^2)^p$ converges. Check my work. 
Given series
$$
\sum\limits_{n=1}^\infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.

To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,
\begin{eqnarray}
\int\limits_1^\infty x(1+x^2)^pdx &=& \lim\limits_{b\to\infty} \int\limits_1^b x(1+x^2)^pdx\\
&=& \lim\limits_{b\to\infty} \left[\dfrac{1}{2(p+1)}\left((1+b^2)^{p+1}-2^{p+1}\right)\right]
\end{eqnarray}

*

*If $p=1$, the integral is divergent,


*If $p>-1$, the integral is divergent,


*If $p<-1$, the integral is convergent to $-\dfrac{1}{p+1}2^p$,

So, I conclude the series is convergent while $p<-1$.
This answer is correct or incorrect?


Alternative:
If $p\ge -1$, then $(1+n^2)^p \ge (1+n^2)^{-1}$
$$\sum_{n=1}^\infty n(1+n^2)^p \ge\sum_{n=1}^\infty \frac{n}{1+n^2} \ge \sum_{n=1}^\infty \frac{n}{2n^2}= \sum_{n=1}^\infty \frac{1}{2n}$$
Hence it diverges. 
If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} \ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$
$$\sum_{n=1}^\infty \frac{n}{(1+n^2)^{-p}} \le \sum_{n=1}^\infty \frac{n}{n^{-2p}}= \sum_{n=1}^\infty \frac{1}{n^{-2p-1}}$$
Hence, by $p$-series and comparison test, it converges. 
