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Determine whether the series $\displaystyle \sum\frac{n^{n+1}}{e^n\left(n+1\right)!}$ is convergent or divergent. I can't use the integral test. I can't use condensation test either, since the series isn't decreasing. Any hints/ideas on how to pursue this exercise?

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  • $\begingroup$ Stirling's approximation to the factorial. $\endgroup$ – A.S. Nov 10 '15 at 21:22
  • $\begingroup$ I'm not acquainted with that, reading on it now. However, I don't think I should be using things that I didn't study so far at my course. $\endgroup$ – MikhaelM Nov 10 '15 at 21:24
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    $\begingroup$ Can you use the Ratio test? (Hoping that it doesn't give an inconclusive answer) $\endgroup$ – imranfat Nov 10 '15 at 21:56
  • $\begingroup$ From what I can scribble now it seems ratio gives me 1/e, which would mean convergence. Hope I didn't get it wrong. Thanks mate! $\endgroup$ – MikhaelM Nov 10 '15 at 22:09
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    $\begingroup$ Unfortunately, the ratio test yields an inconclusive answer. $\endgroup$ – Aneesh Nov 11 '15 at 15:30
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The series is divergent, I will provide a complete answer below, but before that, I want to make sure you know what the 'Monotone Convergence Theorem' is, as well as one of the definitions of '$e$', as they will be relevant in order to understand the answer that will be posted below. The ratio test yields an inconclusive result, but we can still work to show this series is divergent, albeit it is a bit more tricky than I would have anticipated at first glance of the question.

First of all, we define $e$ as follows:

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

Secondly, I will provide you with a link to a statement of the Monotone Convergence Theorem as well as a proof, in case you are unfamiliar with it.

Now, onto the proof.

We will aim to show that $\lim_{n\to\infty} \LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$ exists and is finite.

Once we have shown this, it follows from the definition of the limit to infinity of a sequence, that the following statement is true for all $n$ sufficiently large and some $k>0$

$\large(\frac{1}{n})(\frac{1}{k}) < \frac{n^{n+1}}{e^n(n+1)!}$

Why? I'll explain below:

Recall the definition of the existence of a finite limit to infinity of a sequence ${a_n}$, namely, $\lim_{n\to\infty} a_n = L \iff \forall \epsilon >0 , \exists N\in \mathbb{N}$ such that $\forall n>N, |a_n - L| < \epsilon$.

Read $\forall$ as 'For all' and $\exists$ as 'There exists'.

Thus, suppose $\lim_{n\to\infty} \LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$ $ = l$

Then $\exists N\in \mathbb{N}$ such that $\forall n>N$ $|\LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$$ - l| < 1$.

Now recall the reverse triangle inequality, namely:

$|a|-|b| < |a+b|$

Therefore, we have :

$\LARGE|\LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}|$ - $|l|$$ < 1$

Let $1 + |l| = k>0$

Then, rearranging, we have:

$|\frac{1}{n}| < k|\frac{n^{n+1}}{e^n(n+1)!}|$

rearranging again and ignoring absolute value signs (as we are dealing with positive sequences) we get:

$\large(\frac{1}{n})(\frac{1}{k}) < \frac{n^{n+1}}{e^n(n+1)!}$ for $n $sufficiently large as promised.

This is desirable, because we already know that $\sum{\frac{1}{n}}$ is divergent, so the left hand side of the inequality above is divergent, and thus by the comparison test for positive series $\sum{\frac{n^{n+1}}{e^n(n+1)!}}$ is divergent.

Now all we need to do is actually show that $\lim_{n\to\infty} \LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$ exists and is finite.

We will do so by the Monotone Convergence Theorem.

We will first show that the sequence we are interested in taking the limit of is bounded below. This is easy, as it is always positive for positive integer values of $n$, hence our lower bound is simply $0$.

We can also show that the sequence we are interested in is monotonic decreasing for n sufficiently large as follows:

First of all, let's simplify the expression $\LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$.

This is simply $\large\frac{e^n(n+1)!}{n^{n+2}} = $$a_n$

We will show that for $n$ sufficiently large $a_n \geq a_{n+1}$

This amounts to showing that for $n$ sufficiently large

$\large\frac{a_n}{a_{n+1}}$$ \geq 1$

On simplifying $\large\frac{a_n}{a_{n+1}}$ we see this is nothing but $\large\frac{1}{e}(1+\frac{1}{n})^n(\frac{(n+1)^3}{n^2})$

As $n$ grows increasingly large, we know by the definition of $e$ that $\large\frac{1}{e}(1+\frac{1}{n})^n$ approaches $1$ and $\large(\frac{(n+1)^3}{n^2})$ tends to $+\infty$ and so $\large\frac{a_n}{a_{n+1}}$ grows larger than $1$ for $n$ sufficiently large.

Thus we have shown that for $n$ sufficiently large $a_n = \LARGE\frac{(\frac{1}{n})}{(\frac{n^{n+1}}{e^n(n+1)!})}$ is monotonic decreasing and bounded below, which means by the Monotone Convergence theorem, it converges to a finite limit.

Thus $\sum{\large\frac{n^{n+1}}{e^n\left(n+1\right)!}}$ is divergent.

If you have any questions please feel free to ask, as I am not sure if this explanation is very clear.

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    $\begingroup$ Thank you! Took a while to understand everything, good thing I have my course notes with me as well. Really clever answer. $\endgroup$ – MikhaelM Nov 11 '15 at 15:59
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    $\begingroup$ No problem, glad the answer was useful. $\endgroup$ – Aneesh Nov 11 '15 at 16:04

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