Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $

Question

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.

Answer 1

You can also use the root test together with the inequality between geometric and arithmetic mean (GM-AM):

$$\sqrt[k]{\frac{k!}{k^k}}= \frac{\sqrt[k]{k!}}{k}\stackrel{GM-AM}{\leq}\frac{\frac{k(k+1)}{2k}}{k}=\frac{k+1}{2k}\stackrel{k\to\infty}{\longrightarrow}\frac 12 <1$$

Answer 2

You can apply the Ratio Test and use $\displaystyle{\lim_{k\to\infty}\left(\frac{k+1}{k}\right)^k=e}$.

Answer 3

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.

Answer 4

Since $$ \mathrm{e}^k = 1 + k + \ldots + \frac{{k^k }}{{k!}} + \ldots \geq \frac{{k^k }}{{k!}}, $$ we have $$ \frac{{k!}}{{k^k }} \leq \mathrm{e}^{ - k} . $$ Thus, $$ \sum\limits_{k = 1}^\infty {\frac{{k!}}{{k^k }}} \le \sum\limits_{k = 1}^\infty {\mathrm{e}^{ - k} } = \frac{1}{{\mathrm{e} - 1}}. $$

Answer 5

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.

Answer 6

$\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}} $.