Decide whether the sum is convergent or divergent

Question

$$\sum^{\infty}_{n=0}\frac{1}{n!}(\frac{n}{e})^n$$

I have tried both ratio test and root test, both results in $1$, which cannot give any conclusion. Also the series itself goes to zero as $n$ goes to infinity and the series is not monotonic increasing.

I ran out of methods... I think the only way left is to solve by comparison test but I have no idea what will be the lower bound...

Please help...

Answer 1

Suppose that $n\geq 1$. Observe that \begin{align*} \log n! & = \log n + \sum\limits_{k = 1}^{n - 1} {\log k} = \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log k\, dt} } \le \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log t\, dt} } \\ & = \log n + \int_1^n {\log t \, dt} = \log n + n\log n - n + 1. \end{align*} Taking the exponential of both sides of the inequality gives $$ n! \le n\left( {\frac{n}{e}} \right)^n e \Leftrightarrow \frac{1}{e}\frac{1}{n} \le \frac{1}{{n!}}\left( {\frac{n}{e}} \right)^n . $$ Consequently, $$ \frac{1}{e}\sum\limits_{n = 1}^\infty {\frac{1}{n}} \le \sum\limits_{n = 1}^\infty {\frac{1}{{n!}}\left( {\frac{n}{e}} \right)^n } , $$ but the series on the left-hand side is the divergent harmonic series, i.e., the series in question must diverge too.

Answer 2

Hint:

As this is a series with positive terms, use Stirling's asymptotic equivalent for $n!$.