Find the domain of convergence of $\sum_{n=1}^{\infty } \frac{(n+x)^n}{n^{(n+x)}} $
I've try ratio test for this series, but it doesn't seem to work at all. One thing I know for sure is that by the necessary condition
$$\sum_{n=1}^{\infty } \frac{(n+x)^n}{n^{(n+x)}} \space \space \text{converges} \longrightarrow \lim_{n \to \infty}\frac{(n+x)^n}{n^{(n+x)}}=0$$
and
$$\frac{(n+x)^n}{n^{(n+x)}}= \frac{(n+x)^n}{n^{n} \times n^x}=\left ( \frac{n+x}{n} \right )^n \times \frac{1}{n^x}= \left ( 1+\frac{x}{n} \right )^n \times \frac{1}{n^x} $$
so the necessary condition above becomes $$0=\lim_{n \to \infty}\frac{(n+x)^n}{n^{(n+x)}}=\lim_{n \to \infty} \left ( 1+\frac{x}{n} \right )^n \times \frac{1}{n^x}=e^x \times\lim_{n \to \infty} \frac{1}{n^x} $$
hence I think $x$ must $> 0$ to make sure that $\frac{1}{n^x} $ approaches to $0$ as $n\longrightarrow \infty$
Any recommendation please?
As noticed the series is not in the form of a power series therefore it is meaningless look for the "radius of convergence".
What we can do is study the convergence of the series related to the value of $x$ and since
$$\frac{(n+x)^n}{n^{(n+x)}}= \left ( 1+\frac{x}{n} \right )^n \cdot \frac{1}{n^x}\sim \frac{e^x}{n^x}$$
by limit comparison test with $\sum \frac{1}{n^x}$ we can conclude that the series converges for $x>1$ and diverges otherwise.