Test the convergence of the series $\sum\limits_nn^p(\sqrt{n+1}-2\sqrt n + \sqrt{n-1})$ I am trying to test the convergence of the series
$$\sum_{n=1}^\infty n^p(\sqrt{n+1}-2\sqrt n + \sqrt{n-1})$$
$p$ is a fixed real number.
I tried the ratio test and the integral test without success. Now, I am stuck with the quantity between the round brackets. I have no idea on how to treat it in order to obtain something useful to work with.
 A: 
Tool: For every fixed $a$, when $x\to0$, $(1+x)^a=1+ax+\frac12a(a-1)x^2+o(x^2)$. In particular, the case $a=\frac12$ yields $\sqrt{1+x}=1+\frac12x-\frac18x^2+o(x^2)$.

For every fixed $c$, $$\sqrt{n+c}=\sqrt{n}\cdot\sqrt{1+\frac{c}n}=\sqrt{n}\cdot\left(1+\frac{c}{2n}-\frac{c^2}{8n^2}+o\left(\frac1{n^2}\right)\right),$$ hence $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}=\sqrt{n}\cdot\left(1+\frac1{2n}-\frac1{8n^2}-2+1-\frac1{2n}-\frac1{8n^2}+o\left(\frac1{n^2}\right)\right),$$ that is, $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\sim\sqrt{n}\cdot\frac{-1}{4n^2}=\frac{-1}{4n^{3/2}}.$$ Can you finish this?
A: In essence the problem is to determine the order of 
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}$$
Tring the ususal trick of multiplying by the conjuguate we get
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}
=\frac{2\sqrt{n^2-1}-2n}{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}$$ then doing the same again we get
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}
=\frac{-2}{(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)}$$
Now the order of $(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)$ is $n^{\frac{3}{2}}$ as can be easily verified, since 
$$\frac{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}{\sqrt{n}}\rightarrow 4$$  and 
$$\frac{\sqrt{n^2-1}+n}{n}\rightarrow 2$$
So if we compare our series with the series $\sum n^{p-\frac{3}{2}}$
 we see that we get a finite ratio. Thus the criteria for convergence is 
$$p-\frac{3}{2}<-1$$
or $$p<\frac{1}{2}$$
