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I'm doing a homework problem - it asks us to show whether the series: $$\sum_{n=1}^{\infty}\frac{n^n}{n!}$$ converges or diverges. I looked at a graph of the sequence component $\big(\frac{n^n}{n!} \big)$ and saw it continued to increase. I then considered the sequence: $$\sum_{n=1}^{\infty}\frac{1}{n}$$ which diverges by the P test.

But, $$\frac{1}{n}\leq \frac{n^n}{n!},~ \forall~n\geq1$$ which would then mean that the first series I showed diverges by the comparison test.

My problem is that this seems too simple? Can I compare one series to any series or does the comparison series have to meet some certain requirements (besides those I've addressed).

Cheers.

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    $\begingroup$ Looks fine to me. $\endgroup$ – Jacky Chong Oct 12 '16 at 6:15
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    $\begingroup$ You could also do comparison test with the divergent series $\sum_{n=1}^\infty 1$. $\endgroup$ – angryavian Oct 12 '16 at 6:16
  • $\begingroup$ That seems so simple, so I do not need to compare a series with changing exponents and factorials to a similar one? $\endgroup$ – Wharf Rat Oct 12 '16 at 6:17
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    $\begingroup$ Is the necessary condition for convergence ($n^n/n!\to 0$) satisfied here? $\endgroup$ – A.Γ. Oct 12 '16 at 6:20
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Your reasoning is fine. In fact, this can be shown to diverge with the simplest / most intuitive test there is: If $ \ \displaystyle \sum_{n=1}^\infty a_n$ is a convergent series, it is a necessary condition that $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$. In this scenario, $n^n \geq n!$ for all $n \geq 1$, implying $\displaystyle \frac{n^n}{n!} \geq 1$ for all such $n$, so this necessary condition does not hold.

In short, your observation here:

I looked at a graph of the sequence component $\big(\frac{n^n}{n!} \big)$ and saw it continued to increase.

was all you needed.

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If there series converges then $n^n/n! \to 0$, but if you use Stirling's approximation: $n! \approx \sqrt{2\pi n} \cdot \left(\frac{n}{e}\right)^n$

$$\lim_{n} \frac{n^n}{n!} = \lim_n \frac{n^n}{\sqrt{2 \pi n} \cdot\dfrac{n^n}{e^n}} = \lim_n \frac{e^n}{\sqrt{2 \pi n}} = \frac{1}{\sqrt{2\pi}} \lim_n \frac{e^n}{\sqrt{n}} = \frac{1}{\sqrt{2\pi }} \lim_n \left(\frac{e^{2n}}{n}\right)^{\frac{1}{2}} \to \infty$$

Everything said is perfectly fine, this is just an alternative. The use of Stirling's approximation when limits involve factorials is very useful.

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    $\begingroup$ No need for Stirling here, we have $$\frac{n^n}{n!}=\frac{n\cdot n\cdots n}{1\cdot2\cdots(n-1)\cdot n}>\frac{n\cdot 2\cdots n}{1\cdot2\cdots(n-1)\cdot n}=n$$ $\endgroup$ – AD. Oct 12 '16 at 8:04
  • $\begingroup$ Oh geez. I said I was just providing an alternative. $\endgroup$ – Faraad Armwood Oct 12 '16 at 8:04

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