For what values of $x$ does the series converge: $\sum \limits_{n=1}^{\infty} \frac{x^n}{n^n}$? For what values of $x$ do the following series converge or diverge
$$\sum \limits_{n=1}^{\infty} \frac{x^n}{n^n}$$
I tried to solve this using the ratio test where the series converge when
$$\lim \limits_{n \to \infty} \frac{x^{n+1}n^n}{(n+1)^{n+1}x^n} <1$$
$$\lim \limits_{n \to \infty} \frac{xn^n}{(n+1)^{n+1}} <1$$
but then I am not sure what to do next.
Please give me some ideas or hints on how to solve this question, thanks to anybody who helps.
 A: Use test comparison against geometric series, by noting that for any $n>3|x|$ it holds
$$\frac{|x|^n}{n^n}\leq \left(\frac13\right)^n$$
A: Hints
Upper-bound by a geometric series with ratio between 0 and 1 ... (hover mouse to see more)

 That series will be convergent, and then yours will be assured to converge too because of the direct comparison test.

... (and then some more) ...

 To obtain that convergent geometric series, you may need to remove some initial terms.


Detailed answer
I assume you ask about pointwise convergence, that is, convergence of the numeric series corresponding to a fixed value of $x$. I also assume that $x$ is complex in general.
Given $x$, let $n_0$ be an integer greater than $|x|$, and let $a = |x|/n_0$. Then $0<a<1$, and $|x|/n \leq a$ for all $n \geq n_0$. So, removing the $n_0-1$ initial terms (which doesn't affect convergence),
$$
\left| \sum_{n=n_0}^\infty \frac{x^n}{n^n} \right|
\leq \sum_{n=n_0}^\infty \frac{|x|^n}{n^n}
\leq \sum_{n=n_0}^\infty a^n.
$$
The latter is a geometric series, which is absolutely convergent because $0 < a <1$. So, applying the direct comparison test it stems that for any complex $x$, the original series is absolutely convergent.
A: Hint: $$\dfrac{n^n}{(n+1)^{n+1}} = \dfrac{1}{n+1} \left( \dfrac{n}{n+1}\right)^n < \ldots$$
A: For the ratio test you have to use absolute value as you're not said what $\;x\;$ , and this also works in the complex case. But in this case the n-th root test is better, I believe::
$$\sqrt[n]{\frac{|x|^n}{n^n}}=\frac{|x|}n\xrightarrow[n\to\infty]{}0$$
and the above is true for all $\;x\in\Bbb C\;$ , so the series converges for all the complex.
A: Staring from the last step you have shown we have
$$\lim \limits_{n \to \infty} \frac{xn^n}{(n+1)^{n+1}} <1.$$
But 
$$\lim \limits_{n \to \infty} \frac{xn^n}{(n+1)^{n+1}} =\lim_{n \to \infty} \frac{x}{\left(1+\frac{1}{n}\right)^n(n+1)}.$$
As $n \to \infty$, the limit $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e.$ Thus the limit
$$\lim_{n \to \infty} \frac{x}{\left(1+\frac{1}{n}\right)^n(n+1)}=0 \quad \forall x.$$
A: Note that
$$
n^n \ge n! \implies \frac{x^n}{n^n} \le \frac{x^n}{n!}
$$
and use the comparison test against the series formulation of $e^x$

Note the above only works for positive $x$, however consider showing that the series is absolutely convergent for any $x$ using the above inequalities and thus it is convergent for any $x$.
