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$\sum_{n=0}^\infty (-1)^n \frac{n!}{n^n}$

Wolfram says it converges by ratio test, however L = 1 when i execute the test. am I doing something wrong?

$L = \frac{n}{n+1}$

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  • $\begingroup$ please write down what ever you have done with ratio test in detail.. $\endgroup$ – user87543 Dec 9 '13 at 10:07
  • $\begingroup$ You flubbed up something. I get L = lim $(\frac{n}{n+1})^n=e^{-1}$. $\endgroup$ – David H Dec 9 '13 at 10:11
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Hint: Stirling's formula gives you

$$n\,!\sim \sqrt{2\pi n}\,\left({n \over {\rm e}}\right)^n$$


Alternately, if $a_n=(-1)^n\frac{n!}{n^n}$, then the ratio test runs like this:

$$\left|\frac {a_{n+1} }{a_n}\right|= \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!}=(n+1) \cdot \left( \frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}=\left( \frac{n}{n+1}\right)^n \underset{n \to \infty}\longrightarrow \frac{1}{e} < 1$$


For the last limit in the ratio test, I use the fact that $\left( 1 + \frac{1}{n}\right)^n \underset{n \to \infty}\longrightarrow e$. If you don't know this, you can prove it, using $\log \left(1+\frac{1}{n}\right) = \frac{1}{n} + O\left(\frac{1}{n^2}\right)$:

$$\left(1+ \frac{1}{n}\right)^n=\exp \left(n \log \left(1+\frac{1}{n}\right)\right)=\exp \left(1+O\left(\frac{1}{n}\right)\right) \underset{n \to \infty}\longrightarrow e$$


If you don't know either this limit or the big-O notation, you can still do something: define

$$u_n=\left(1+\frac{1}{n}\right)^n$$ $$v_n=\left(1+\frac{1}{n}\right)^{n+1}$$

Then you can prove that $u_n$ is increasing, $v_n$ is decreasing (not difficult, but rather tedious algebraic manipulations), and of course $u_n < v_n$ since $v_n=\left(1+\frac{1}{n}\right) u_n$. Thus $u_n$ is convergent (this convergence follows from the fact that $u_n$ is increasing and bounded, by $v_1$), and its limit must be greater than $u_1=2$. Thus the limit in the ratio test is less than $1 \over 2$.

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  • $\begingroup$ I am afraid he may not be familiar with stirling's formula assuming that he is just started real analysis... I don't know if this is the right way to help one who has just started learning real analysis... $\endgroup$ – user87543 Dec 9 '13 at 10:09
  • $\begingroup$ Ok, added a more detailed approach $\endgroup$ – Jean-Claude Arbaut Dec 9 '13 at 10:14
  • $\begingroup$ I am sorry if that has bothered you much.. I was just expressing my intention.. Thank you for the explanation :) $\endgroup$ – user87543 Dec 9 '13 at 10:16
  • $\begingroup$ No worry, you were right, Stirling should not be expected from beginners. It's just that when I see $\frac{n!}{n^n}$, it's like a reflex ;-) $\endgroup$ – Jean-Claude Arbaut Dec 9 '13 at 10:19
  • $\begingroup$ OH i didn't add the plus one on the bottom $n^n$ just got $n^(n+1)$ $\endgroup$ – John Dec 9 '13 at 10:36
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You could also consider the alternating series test: $$\sum \limits_{i=1}^{\infty} (-1)^{n}a_{n} $$ converges if the sequence $a_{n}$ is monotonically decreasing and null.

In your sum, the sequence satisfies this, so you do have convergence.

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