Find x so that $\sum^{\infty}_{n=1}{\frac{1}{(x-1)^n\sqrt{n+5}}}$ converges I want to get find the $x$ values that make this series converge (simply and absolutely):
$$\sum^{\infty}_{n=1}{\frac{1}{(x-1)^n\sqrt{n+5}}}$$
First I study the absolute convergence by using an absolute value:
$$\left|{{\frac{1}{(x-1)^n\sqrt{n+5}}}}\right|$$
We get a positive series so I can now use the root test (this is because we have a root of $n$ so I would like to get rid of it) and I get:
$$\left|{{\frac{1}{(x-1)\sqrt[2n]{n+5}}}}\right|=\dots$$
I don't really know what to do here. Also, I'm not sure if I have to use the Cauchy criterion for series every time I want to test the convergence of a series. Any hints?
 A: By root test, since for any polynomial $p(n)$ we have that $\sqrt[n]{p(n)} \to 1$, we obtain
$$\left|{{\frac{1}{x-1}}}\right|{{\frac{1}{\left(\sqrt[n]{(n+5)}\right)^\frac12}}}\to \left|{{\frac{1}{x-1}}}\right|$$
therefore the series converges for
$$\left|{{\frac{1}{x-1}}}\right|<1 \iff|x-1|>1 \iff x-1<-1 \lor x-1>1 \iff x<0 \lor x>2$$
More in detail we have that


*

*for $x>2 \implies x-1>1$


$$\sum^{\infty}_{n=1}{\frac{1}{(x-1)^n\sqrt{n+5}}}$$
which converges for example by limit comparison test with $\sum \frac 1 {n^2}$.


*

*for $x=2$


$$\sum^{\infty}_{n=1}{\frac{1}{\sqrt{n+5}}}$$
which diverges.


*

*for $1<x<2 \implies 0<x-1<1$ by $y=\frac 1{x-1}>1$


$$\sum^{\infty}_{n=1}{\frac{y^n}{\sqrt{n+5}}}$$
which diverges.


*

*for $x=0$


$$\sum^{\infty}_{n=1}{\frac{(-1)^n}{\sqrt{n+5}}}$$
which converges by alternating series test.


*

*for $0<x<1 \implies 0<1-x<1$ by $y=\frac 1{1-x}>1$


$$\sum^{\infty}_{n=1}{\frac{y^n(-1)^n}{\sqrt{n+5}}}$$
which diverges.


*

*for $x<0 \implies 1-x>1$


$$\sum^{\infty}_{n=1}{\frac{(-1)^n}{(1-x)^n\sqrt{n+5}}}$$
which converges by alternating series test.
A: Note that $\sqrt[2n]{n+5}$ = $\left(\sqrt[n]{n+5}\right)^{1/2}.$
A: Hint:
The ratio test might be easier: what is
$$\frac1{|x-1|}\sqrt{\frac{n+6}{n+5}}\,?$$
