Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
 = \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
 A: You can also use the root test together with the inequality between geometric and arithmetic mean (GM-AM):
$$\sqrt[k]{\frac{k!}{k^k}}= \frac{\sqrt[k]{k!}}{k}\stackrel{GM-AM}{\leq}\frac{\frac{k(k+1)}{2k}}{k}=\frac{k+1}{2k}\stackrel{k\to\infty}{\longrightarrow}\frac 12 <1$$
A: You can apply the Ratio Test and use $\displaystyle{\lim_{k\to\infty}\left(\frac{k+1}{k}\right)^k=e}$.
A: Here is another elementary solution. Using the inequality $x(a-x) \leq \frac{a^2}{4}$ for $0 \leq x \leq a$, we get
$$ \frac{k!}{k^k} = \frac{(k-1)!}{k^{k-1}} = \prod_{j=1}^{k-1} \frac{\sqrt{j(k-j)}}{k} \leq \frac{1}{2^{k-1}}. $$
Therefore the series converges by the Direct Comparison Test.
A: You can also use Stirling's formula for the asymptotics of $k!$ as $k\to\infty$ to bound it:
$$k! = \mathcal{O}\left(k^{k+\frac{1}{2}} e^{-k}\right),$$
which you can derive from a Laplace-type expansion of $k!=\int_0^\infty e^{k \log x} e^{-x} \,dx$ for large $k$. Hence, your summand is $\mathcal{O}\left(\sqrt{k}\,e^{-k}\right)$, meaning that its sum over $\mathbb{N}$ converges.
A: $\begin{array}\\
f(n)
&=\dfrac{n!}{n^n}\\
&=\dfrac{\prod_{k=1}^n k}{n^n}\\
&=\prod_{k=1}^n (k/n)\\
&=\prod_{k=0}^{n-1} ((n-k)/n)\\
&=\prod_{k=0}^{n-1} (1-k/n)\\
g(n)
&=\ln(f(n))\\
&=\sum_{k=0}^{n-1} \ln(1-k/n)\\
&<\sum_{k=0}^{n-1} -k/n
\qquad \ln(1-x) < -x\\
&=-\dfrac{n(n-1)}{2n}\\
&=-\dfrac{n-1}{2}\\
&=\dfrac12=\dfrac{n}{2}\\
\text{so}\\
f(n)
&\le e^{-(n-1)/2}\\
&=e^{1/2}(e^{-1/2})^n\\
\text{so}\\
\end{array}
$
$\sum_{n=1}^{\infty} f(n)
\le \dfrac{e^{1/2}}{1-e^{-1/2}}
$.
