Ok so I'm working on this problem but have problem understanding it. I got the right answer but I made an assumption to get there. I need help proving that assumption.

Here is the problem

Find the values of p for which the series in convergent

$\displaystyle \sum_{n=1}^{\infty} n(1+n^2)^p$

Ok so I used the integral test because f(x) = $\displaystyle x(1+ x^2)^p$ is continous, positive, and decreasing(this is where I made the assumption that it was decreasing).

$\displaystyle \int_{1}^{\infty} x(1+x^2)^{p}dx = \lim_{t \to \infty}\bigg[\frac{(1+x^2)^{p+1}}{2p+2}\bigg]_1^t = \lim_{t \to \infty}\bigg(\frac{1}{2}\bigg)\bigg(\frac{(1+t^2)^ {p+1}}{p+1}\bigg) - \frac{2^p}{p+1}$

Thus the limit exist when $\displaystyle p + 1 < 0$ so the series converges when p< -1.

Is that right? Also how do I show that the function is decreasing(the assumption I need to prove). I took the 1st derivative and equalled it to less than 0 to show me where it was decreasing but ran into problems.

This is what I get stuck on with the derivative

$\displaystyle x^2(1+2p)< -1$

Thanks