Series converge or diverge : $\sum^\infty_{n=1}n(1+n^2)^p$ , $p\in\mathbb{R}$? 
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*$\sum^\infty_{n=1}n(1+n^2)^p$ , $p\in\mathbb{R}$


I tried to compare it with other known sequencies but i couldn't find the right one.
I also tried to solve it using Mathematical Induction (for p=-1) but it led me nowhere.
Any suggestions about what should I do?

Note that $\dfrac {(1+n^2)^p} {n^{2p}} = \left(1 + \dfrac 1 {n^2} \right) ^p \to 1 \ \forall p \in \Bbb R$, so you may use the limit comparison test to deduce that your series exhibits the same behaviour as the series $\sum n \cdot n^{2p} = \sum \dfrac 1 {n^{-(2p+1)}}$ (the generalized harmonic series), which is known to be convergent for $-(2p+1) > 1$ and divergent for $-(2p+1) \le 1$. This gives convergence for $p<-1$ and divergence for $p \ge -1$.
