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.
On
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 $\beta>1$.
$$2-\alpha+\beta/2+0\implies \alpha=2+\beta/2,\quad\text{and}\\ \beta>1\implies \alpha>5/2.$$
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. :) )