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I'm trying to show that $$\sum_{n=1}^\infty \left(\frac{1}{n!}\right)\left(\frac{n}{e}\right)^n$$ diverges by the Corollary of Raabe's Test.

Corollary of Raabe's Test: Let $X=\{x_n\}$ be a sequence of nonzero real numbers and let $a=\lim_{n \to \infty} (n(1-\frac{|x_n+1|}{x_n}))$, whenever this limit exists. Then the series is absolutely convergent when $a>1$ and is divergent when $a<1$.

What I find:

$$a= \lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\infty \cdot 0$$ (which is indeterminate)

$$\lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\lim_{n \to \infty} \frac{1-\frac{1}{e}(1+\frac{1}{n})^n)}{\frac{1}{n}}$$

Applying L'hospital's rule, I get: $$\lim_{n \to \infty} \frac{-\frac{n}{e}(1+\frac{1}{n})^{n-1}(-\frac{1}{n^2})}{-\frac{1}{n^2}}=\lim_{n \to \infty} (-\frac{n}{e}(1+\frac{1}{n})^{n-1})$$

Help!!! I do not know what to do next.

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    $\begingroup$ Do you know Stirling’s Formula? I’d recommend using it, if so $\endgroup$ – Clayton Jan 31 '18 at 12:59
  • $\begingroup$ @Gie you need to consider second order terms as shown by Crostul and me. As an alternative there are more efficient method but I guess you know them. $\endgroup$ – user Jan 31 '18 at 15:15
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The limit is $a=1/2$.

Start by the approximation

$$\ln \left( \frac 1e (1+x)^{1/x} \right) = -1 + \frac 1x \ln (1+x) = -1 + 1-\frac 12 x +o(x) = -\frac 12 x +o(x) $$

Plugging $x= \frac 1n$ you get

$$1- \frac 1e \left( 1+ \frac 1n \right)^n = 1- e^{\ln \left( \frac 1e (1+\frac 1n)^{n} \right)} = 1- e^{-\frac{1}{2n} + o( 1/n)}$$

Now, again, using the limit $$\lim_{x \to 0} \frac{1-e^x}{x} = -1$$

we have $$a= \lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{1/n} = \lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{-\frac{1}{2n} + o( 1/n)} \cdot \frac{-\frac{1}{2n} + o( 1/n)}{1/n} = -1 \cdot \left( -\frac{1}{2} \right) = \frac{1}{2}$$

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  • $\begingroup$ It's my first time seeing this, may I know how this: $ -1 + \frac {1}{x} \ln (1+x) = -1 + 1-\frac {1}{2} x +o(x) $ happens? $\endgroup$ – Gie Feb 1 '18 at 5:11
  • $\begingroup$ @Gie Simply Taylor expand $\ln (1+x) = x- \frac 12 x^2 + o(x^2)$, then divide by $x$ and subtract $1$. Are you familiar with Taylor expansion? $\endgroup$ – Crostul Feb 1 '18 at 7:35
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Using the elementary inequality $n! < (n/e)^{n+1}$, each term $>e/n$ so the sum diverges.

If we use Stirling, each term is about $1/\sqrt{2\pi n}$ and this gives the true rate of divergence.

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  • $\begingroup$ Maybe the OP was about Raabe's Test. $\endgroup$ – user Jan 31 '18 at 15:09
  • $\begingroup$ OP wrote that was trying to show sum diverges. Raabe's test was an attempt. It was not stated that the test was required. $\endgroup$ – marty cohen Feb 6 '18 at 21:03
  • $\begingroup$ not originally GuyFsone edited the OP $\endgroup$ – user Feb 6 '18 at 21:05
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We have $m=\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)$ (this is a telescopic product), hence $$ n! = \prod_{m=2}^{n}m=\prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^{n-1}}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}. $$ Now we may exploit a classical result: the sequence $\left\{\left(1+\frac{1}{k}\right)^{k+1/2}\right\}_{k\geq 1}$ is decreasing and convergent to $e$. This can be proved through the Hermite-Hadamard inequality, for instance. It implies: $$ n! = \frac{n^{n-1/2}}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k+1/2}} \leq \left(\frac{n}{e}\right)^n e\sqrt{n}$$ and the given series is divergent by comparison with $\sum_{n\geq 2}\frac{1}{e\sqrt{n}}$.

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By Stirling approximation we have $$\lim_{n\to\infty}\frac{\sqrt{2\pi n}}{n!}\left(\frac{n}{e}\right)^n = 1$$

Hence $$\frac{1}{n!}\left(\frac{n}{e}\right)^n\sim \frac{1}{\sqrt{2\pi n}}$$

Therefore the divergence follows since the series $$\sum \frac{1}{\sqrt{2\pi n}}$$

diverges as well

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  • $\begingroup$ Maybe the OP was about Raabe's test. $\endgroup$ – user Jan 31 '18 at 15:09
  • $\begingroup$ @gimusi Raabe deal with first oder approximation which is the same as computing the limit am I correct or not? $\endgroup$ – Guy Fsone Jan 31 '18 at 15:11
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    $\begingroup$ The OP is asking to show the divergency by the Corollary of Raabe's Test, so I think that your answer is completely off topic here. $\endgroup$ – user Jan 31 '18 at 15:14
  • $\begingroup$ Why did you edited the original OP changing the original request? It is a very deprecable and uncorrect behaviour! $\endgroup$ – user Feb 6 '18 at 21:07
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Note that

$$\frac{|x_n+1|}{x_n}=\frac{(n+1)^{n+1}}{e^{n+1(n+1)!}}\frac{e^nn!}{n^n}=\frac1e \left(1+\frac1n\right)^n\to 1 $$

note also that

$$ \left(1+\frac1n\right) ^n= e^{n\log\left(1+\frac1n\right)}= e^{n\left( \frac1n-\frac1{2n^2}+o(n^{-2})\right)} =e^{ \left(1 -\frac1{2n}+o(n^{-1})\right)}=e\left( 1 -\frac1{2n}+o(n^{-1}) \right) $$

thus

$$\frac{|x_n+1|}{x_n}=\frac1e \left(1+\frac1n\right)^n= 1 -\frac1{2n}+o(n^{-1}) $$

and

$$a=n\left(1- \frac{|x_n+1|}{x_n} \right)=n\left(1- 1 +\frac1{2n}+o(n^{-1}) \right)=\frac12+o(1)\to\frac12<1$$

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