Does $\sum \dfrac{n!}{n^n}$ converge? Intuitively I think it converges but I am not sure what test can be used to show that.
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$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!} =\frac{n^n}{(n+1)^n}=\frac1{\left(1+\frac1n\right)^n}\to\frac1e$$ Remark: While $\lim_{n\to \infty} \left(1+\frac1n\right)^n$ it is often used as a definition of $e$, we don't even need convergence here. It suffices to have $\left(1+\frac1n\right)^n\ge 2$ (for $n\ge 1$), which follows from the Bernoulli inequality or simply from the first two summands of the binomial expansion. |
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If you don't know stirlings formula, by the inequality of arithmetic and geometric means we have, $$(\prod_{k=1}^n k)^{\frac{1}{n}}\leq\frac{1}{n}\sum_{k=1}^n k=\frac{n+1}{2}$$ $$\text{ So we have}$$ $$(n!)^\frac{1}{n}\leq\frac{n+1}{2}$$ $$\frac{n!}{(n+1)^n}\leq\frac{1}{2^n}$$ $$\frac{(n+1)!}{(n+1)^n}\leq\frac{(n+1)}{2^n}$$ $$\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}\leq\sum_{n=0}^\infty\frac{(n+1)}{2^n}=4$$ So, $\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}$, converges |
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From Stirling's formula, we have $$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}e \right)^n$$ Hence, $$\dfrac{n!}{n^n} \sim \dfrac{\sqrt{2 \pi n}}{e^n} \implies \left(\dfrac{n!}{n^n} \right)^{1/n} \sim \dfrac{\sqrt[2n]{2 \pi n}}{e} \to \dfrac1e < 1$$ Hence, $\displaystyle \sum_{n}\dfrac{n!}{n^n}$ converges. |
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$$\frac{n!}{n^n} = \frac{1}{n} . \frac{2}{n} \dots \frac{n}{n} \leq \frac{1}{n} .\frac{2}{n} . 1 \times 1 \dots \times 1 = \frac{2}{n^2}. $$ |
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