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How would you go about proving the infinite series $\frac{n!}{n^n}$ converges or diverges?

alexwlchan
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Dr. Luca
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2 Answers2

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Since

$$\lim_{n\to \infty} \dfrac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \lim_{n\to \infty} \left(\frac{n}{n+1}\right)^n = \lim_{n\to \infty} \frac{1}{(1 + \frac{1}{n})^n} = \frac{1}{e}< 1,$$

your series converges by the ratio test.

kobe
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We first use the zero test that states:

$$\sum{a_n} \ \text{diverges if} \lim_{x\to\infty}a_n\neq0.$$

So we test the series and get that

$$\lim_{x\to\infty}\frac{n!}{n^n}=0.$$

By this result, we fail to show that the series diverges, however it is inconclusive. Therefore, we must use another test.

I did this by the ratio test. The test states as follows:

$$\sum{a_n} \ \text{with} \ \lim_{x\to\infty} \frac{|a_{n+1}|}{|a_n|} \ \text{where it converges if} \ L<1.$$

So, if we work the series out by this test, we get

$$\lim_{n\to \infty} \dfrac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \lim_{n\to \infty} \left(\frac{n}{n+1}\right)^n = \lim_{n\to \infty} \frac{1}{(1 + \frac{1}{n})^n} = \frac{1}{e}< 1.$$

In conclusion, since $L<1$, $\sum\frac{n!}{n^n}$ converges.

Nathan Smith
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Julian Rachman
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    Thanks for your answer nicely done - makes a lot of sense! – Dr. Luca Mar 07 '15 at 19:52
  • May I ask, "why the downvotes?" – Julian Rachman Mar 07 '15 at 20:31
  • @JulianRachman "By this result, we can see that the series does not diverge, however it is still inconclusive that the series converges." This is not right at all. The theorem you used gives sufficient conditions for convergence, not necessary. And "converges xor diverges" is a tautology, so the statement itself does not make sense. (edit: it's fixed now, though only the ratio test is necessary, undid -1). – Eric Thoma Mar 07 '15 at 20:32
  • @Eric it was edited – Julian Rachman Mar 07 '15 at 20:33