<s>[INST] Convergence of $\sum_{n=1}^\infty n^s(\sqrt {n+1} - 2 \sqrt n + \sqrt {n-1})$ Convergence of $$\sum_{n=1}^\infty n^s(\sqrt {n+1} - 2 \sqrt n + \sqrt {n-1})$$
Attempt: $\sum_{n=1}^\infty n^s(\sqrt {n+1} - 2 \sqrt n + \sqrt {n-1}) \sim \sum_{n=1}^\infty n^s( \sqrt n )$ 
As $\lim_{n\rightarrow \infty} \dfrac {n^s(\sqrt {n+1} - 2 \sqrt n + \sqrt {n-1}) }{n^s( \sqrt n )} = 0$
$ \implies \sum_{n=1}^\infty n^s(\sqrt {n+1} - 2 \sqrt n + \sqrt {n-1})$ converges whenever $\sum_{n=1}^\infty n^s( \sqrt n ) = \sum_{n=1}^\infty n^{s+\frac {1}{2}}$ converges.
This happens when $s + \dfrac {1}{2}<-1$ or $s < \dfrac {-3}{2}$
Is my attempt correct? 
Thank you for reading through!
 [/INST] Consider that
$$
\sqrt{n+1}+\sqrt{n-1}=\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}\right) 
$$
Also, we have that as $x$ goes to $0$:
$$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+o(x^2)$$
and
$$\sqrt{1-x}=1-\frac{1}{2}x-\frac{1}{8}x^2+o(x^2).$$
Then, setting $x=\frac{1}{n}$, the sum is asymptotic exactly to 
$$\sum_{n\ge 1}n^{s+\frac{1}{2}}\left(2-\frac{1}{4n^2}-2\right)=-\frac{1}{4}\sum_{n\ge 1}n^{s-\frac{3}{2}}=-\frac{1}{4} \zeta\left(\frac{3}{2}-s\right),$$ 
which converges iff $s$ is smaller than $\frac{1}{2}$.
</s>