Interval of convergence power series I'm having serious problems finding the interval of convergence of the following series:
$$\sum_1^\infty \frac{(x+5)^{n^2}}{(n+1)^{n}} $$
I'll denote 
$$a_{n}=\frac{1}{(n+1)^{n}}$$
$$u_{n}=\frac{(x+5)^{n^{2}}}{(n+1)^{n}}$$
What I have tried:


*

*$$\lim_{n \to \infty} {\frac{|u_{n+1}|}{|u_{n}|}} = \lim_{n \to \infty} {\frac{|(x+5)^{2n+1}(n+1)^{n}|}{|(n+2)^{n+1}|}}< 1$$
Which I failed to evaluate and find the interval.

*$$ Radius \  of \ convergence = \lim_{n \to \infty} {\frac{|a_{n}|}{|a_{n+1}|}} = \lim_{n \to \infty} {\frac{|(n+2)^{n+1}|}{|(n+1)^{n}|}} = \infty $$
This should say that the interval of convergence is $$x \in(-\infty,\infty)$$
But by experimenting with online calculators I've found out that the interval is most likely $$x \in [-6, -4]$$

*I've also tried 
$$ Radius \  of \ convergence = \lim_{n \to \infty} {\frac{1}{\sqrt[n]{|a_{n}|}}} = \infty $$
Which is basically the result I came up with in second test


What should I do? Could you help me please?
 A: (Long comment)
In your method 1 you have
$$
\frac{|u_{n+1}|}{|u_n|} = |x+5|^n \frac{(n+1)^n}{(n+2)^{n+1}}
= |x+5|^n \frac{1}{n+2} \left(1-\frac{1}{n+2}\right)^n
$$
and now you should be able to compute the limit.
(The last factor goes to $e^{-1}$.)
A: (1). If $|x+5|\leq 1$ then $|(x+5)^{n^2}/(n+1)^n|\leq 1/(n+1)^n,$ which is less than $1/2^n$ when $n\geq 1.$ 
If $|x+5|>1$ then $|x+5|^n/(n+1)\to \infty$ as $n\to \infty,$ so $(\;|x+5|^n/(n+1)\;)^n=|(x+5)^{n^2}/(n+1)^n|$ also $\to \infty.$
(2). In your attempt at  the Ratio Test you took $a_n=1/(n+1)^n$ as the co-efficient of $x^n,$ but it isn't. It's the co-efficient of $x^{n^2}.$
(3). The Hadamard Radius Formula: Let $S=\lim_{n\to \infty}\sup_{m\geq n}|A_m|^{1/m}.$  If $|z|<1/S$ then $\sum_nA_n z^n$ converges. If $|z|>1/S$ then $\sum_nA_n z^n$ diverges.
Let $A_n=1/(1+\sqrt n\;)^{\sqrt n}\;$ when $n$ is a perfect square, and $A_n=0$ when $n$ is not. Then  your series is $\sum_n A_n(x+5)^n,$  and each $|A_n|^{1/n}$ is either $0$ or $(1+\sqrt n\;)^{-1/\sqrt n},$ so  the Hadamard formula gives $S=1$.
Convergence at $x=-4$ or $x=-6$, when $|x+5|=1,$ are not handled by the formula and have to be examined  separately.
