Testing the convergence of the series $\sum_{n=1}^{+\infty}\frac{n^{n-2}}{e^nn!}$

Question

Question: test the convergence or divergence of the following series : $$\sum_{n=1}^{+\infty}\frac{n^{n-2}}{e^nn!}$$ What I tried: I could not find a conclusion using d'Alembert ratio test, how would we go about doing this using rabbes test? I'm a first year college student.

Answer 1

Use the Stirling approximation $$ n!\sim\sqrt{2\pi n}\frac{n^n}{e^n}, $$ so that the term of the series becomes $$ \sim\frac1{\sqrt{2\pi}n^{5/2}} $$ which would give a converging series.

Answer 2

Here is a convergence test: If $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ are series of positive terms and $\frac{a_{n+1}}{a_n}\leq\frac{b_{n+1}}{b_n}$ holds for all $n\in\mathbb{N}$ then $\sum\limits_{n=1}^{\infty} b_n<\infty$ implies $\sum\limits_{n=1}^{\infty} a_n<\infty$.

If you haven't seen this convergence test before then try to prove it. As a hint, note that:

$a_n=\frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdot...\cdot\frac{a_2}{a_1}\cdot a_1$

Now use this test with $a_n=\frac{n^{n-2}}{e^n n!}$ and $b_n=\frac{1}{n^2}$.

Answer 3

The sequence ${n^n\over e^n n!}$ is decreasing. Indeed $$ {n^n\over e^n n!}{e^{n+1}(n+1)!\over (n+1)^{n+1}}={e\over (1+{1\over n})^n}>1$$ Thus $$a_n={n^n\over e^n n!}{1\over n^2}\le {1^1\over e^11!}{1\over n^2}={1\over en^2}$$

Answer 4

First observe that if $a_n=\dfrac{n^{n-2}}{e^n n!},$ then $$ \frac{a_{n+1}}{a_n}=\frac{\big(1+\frac{1}{n}\big)^n}{e}\cdot \left(\frac{n}{n+1}\right)^2<\left(\frac{n}{n+1}\right)^2 $$ since $\big(1+\frac{1}{n}\big)^n<e$. So $$ a_n=\frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdots\frac{a_2}{a_{1}}\cdot a_1 < \left(\frac{n-1}{n}\right)^2 \left(\frac{n-2}{n-1}\right)^2\cdots \left(\frac{1}{2}\right)^2 a_1=\frac{a_1}{n^2}=b_n $$ Then, according to the comparison test, since $\sum b_n<\infty,\,$ then also $\sum a_n<\infty$.