Determine all real $x$ for which the following series converges: $$\sum_{k=1}^\infty\frac{k^k}{k!}x^k.$$ You may use the fact that $$\lim_{k\to\infty}\frac{k!}{\sqrt{2\pi k}(k/e)^k}=1.$$

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I found that the radius of convergence is $\displaystyle \frac{1}{e}$.Next considered end points. When $\displaystyle x=-\frac{1}{e}$, the series converges by alternating series test. But I was fail to determine it for the case where $\displaystyle x=\frac{1}{e}$. I tried to use comparison test by using the given fact. I guess in this case the series diverges. Can anybody please give me a hint to complete the proof?

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    $\begingroup$ if you substitute $x=1/e$ you get something very similar to the limit expression in the hint, can you use it? $\endgroup$ – gt6989b Jul 20 '15 at 19:28
  • $\begingroup$ Yes. I did it. But I was fail to compare them. $\endgroup$ – Extremal Jul 20 '15 at 19:29
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    $\begingroup$ Let $x=1/e$ and $k$ large. Then $\frac{k^k}{k!}x^k \gt \frac{1}{2}\cdot \frac{1}{\sqrt{2\pi k}}$, so we have divergence, essentially by comparison with $\sum_2 \frac{1}{\sqrt{k}}$. Or more easily use Limit Comparison. By the way, for $-1/e$ you would need to prove alternating, which the limit statement is not enough for. $\endgroup$ – André Nicolas Jul 20 '15 at 19:35

Let $x=1/e$, and let $a_k=\frac{k^k}{k!}(1/e)^k$. We want to show that $\sum a_k$ diverges.

Let $\frac{k!}{\sqrt{2\pi k}(k/e)^k}=c_k$. We are told that $\lim_{k\to\infty} c_k=1$.

With some algebra, we can see that $$a_k=\frac{k^k}{k!}(1/e)^k=\frac{1}{c_k}\cdot \frac{1}{\sqrt{2\pi k}}.$$ Note that the limit of $\frac{a_k}{1/\sqrt{k}}$ as $k\to\infty$ is a non-zero constant. Thus, by the Limit Comparison Test, $\sum_{k=1}^\infty a_k$ diverges.

Remark: We want to show (conditional) convergence at $x=-1/e$, using the alternating series test. So we want to show that the terms are decreasing in absolute value and have limit $0$. That the limit is $0$ follows from the Stirling Formula, but that is not enough to prove monotonicity.

Look instead at the ratio $\frac{b_{k+1}}{b_k}$, where $b_n=\frac{n^n}{n!}(1/e)^n$. After some simplification, we find that $$\frac{b_{k+1}}{b_k}=\frac{(k+1)^{k+1}}{k^k\cdot (k+1)}(1/e)=\frac{1}{e}\left(1+\frac{1}{k}\right)^k.$$

It is a standard fact that $\left(1+\frac{1}{k}\right)^k$ increases monotonically, and has limit $e$. It follows that $\frac{1}{e}\left(1+\frac{1}{k}\right)^k\lt 1$, and therefore $b_{k+1}\lt b_k$.


As a side note, I think it is worth mentioning that by the Lagrange inversion formula we have: $$ \forall x\in\left(-\frac{1}{e},\frac{1}{e}\right)\qquad W(x) = \sum_{n\geq 1}\frac{(-n)^{n-1}}{n!} x^n $$ with $W$ being the Lambert $W$ function, i.e. the inverse function of $x e^x$ in a neighbourhood of the origin. Trivial manipulations then leads to:

$$ \forall x\in\left(-\frac{1}{e},\frac{1}{e}\right)\qquad \sum_{k\geq 1}\frac{k^k}{k!}\,x^k = \frac{-W(-x)}{1+W(-x)}.$$ The convergence in the endpoints may be studied through Stirling's inequality: $$ \sqrt{2\pi k}\left(\frac{k}{e}\right)^k \exp\left(\frac{1}{12k+1}\right)\leq k!\leq \sqrt{2\pi k}\left(\frac{k}{e}\right)^k \exp\left(\frac{1}{12k}\right).$$

  • $\begingroup$ I have the feeling that $\forall x\in\left(-\infty,\frac{1}{e}\right)$ would work. Am I wrong ? $\endgroup$ – Claude Leibovici Jul 21 '15 at 6:10
  • $\begingroup$ @ClaudeLeibovici: a power series cannot converge past its radius of convergence! $\endgroup$ – Jack D'Aurizio Jul 21 '15 at 14:35
  • $\begingroup$ I was considering $ \frac{-W(-x)}{1+W(-x)}$ which seems (to me) to be defined $\forall x\in\left(-\infty,\frac{1}{e}\right)$ $\endgroup$ – Claude Leibovici Jul 22 '15 at 5:28
  • $\begingroup$ @ClaudeLeibovici: oh, sure. That gives that the RHS is indeed an analytic continuation of the LHS. I was just saying that if we take some $x$ such that $|x|>\frac{1}{e}$, there is no hope that the series in the LHS is converging. $\endgroup$ – Jack D'Aurizio Jul 22 '15 at 13:52

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