How to solve this sequence? I have this sequence: $\sum_{n=1}^{\infty} \frac{n^2+n-1}{\sqrt{n^\alpha+n+3}}$
For which values of $\alpha$ does this converge?
I first tried to separate into cases where $\alpha \gt 0$ etc and using the ratio test.. but it seems like it doesn't help.
Suggestions?
Many thanks.
 A: First note that $\sum_{n=0}^\infty 1/x^n$ converges if and only if $n>1$
$$\begin{align}
& \lim_{n\to\infty}{(n^2+n-1)/\sqrt{x^\alpha+x+3}\over 1/n^\beta}\\
= & \lim_{n\to\infty}{n^{2-\beta/2}+n^{1-\beta/2}-n^{-\beta/2}\over\sqrt{n^{\alpha-\beta}+n^{1-\beta}+3n^{-\beta}}}\\
= & \lim_{n\to\infty}n^{2-\alpha+\beta/2}
\end{align}$$
If $2-\alpha+\beta/2=0$, then this limit is $1$ and the given series will converge if and only if $\sum 1/x^\beta$ converges. As noted above this is when $\beta>1$.
$$2-\alpha+\beta/2+0\implies \alpha=2+\beta/2,\quad\text{and}\\
\beta>1\implies \alpha>5/2.$$
A: For large $n$, the numerator is less than $2n^2$, and the denominator is larger than $n^{\alpha / 2}$, so the tail of the sum is dominated by 
$$
2\sum \frac{n^2}{n^{\alpha / 2}} = 2 \sum n^{2 - \alpha/2}
$$
This converges when 
$$
2 - \frac{\alpha}{2} < -1 \\
3 < \frac{\alpha}{2} \\
6 < \alpha.
$$
On the other hand, for large $n$, the numerator is greater than $n^2$ and the denominator less than $2 n ^ \frac{\alpha}{2}$, so the tail of the sum is at least
$$
\frac{1}{2}\sum \frac{n^2}{n^{\alpha / 2}} = \frac{1}{2} \sum n^{2 - \alpha/2}
$$
so the same reasoning shows that the series diverges for $6 \ge \alpha$. 
(My answer differs from @Tim's, so one of us made an algebra mistake. I hope it's him. :) )
