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so I used the root test, but i'm not quite sure if i'm allowed to. I think im performing the operations correctly,a and i keep ending up with $(1)^{\infty}$. So really my question is am i performing the operations wrong or do i have to use a different test?

Stahl
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lawlipop
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5 Answers5

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Hint:

$$\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{1}{\left(1+\frac{1}{n}\right)^n}\xrightarrow[n\to\infty]{}\ldots$$

And yes: it converges.

DonAntonio
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    Why the "-" before the 1/...? – marty cohen Apr 08 '13 at 02:59
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    @DonAntonio yea i too was wondering about the negative? – lawlipop Apr 08 '13 at 02:59
  • Notice the ratio of successive terms converges to $1/e$. – obataku Apr 08 '13 at 03:10
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    That minus after the equality sign must have been an act of god: I've no idea how it got there...well, I do have an idea but putting the blame on god makes me feel better. Thanks. – DonAntonio Apr 08 '13 at 03:20
  • The limit of the above is $\frac 1e \approx 0.36788$, which is the same thing I got in a more roundabout way. But why does WolframAlpha report the answer is 1.87985 (see: http://tinyurl.com/n99hwld) – Jeff Mar 20 '14 at 23:02
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    @Jeff Because one thing is the limit of the sequence in the above answer, and way another is the sum of the series the question asks about... – DonAntonio Mar 21 '14 at 00:17
  • Why does this converge to $\frac{1}{e}$? Where does the $e$ come from? – mavavilj Sep 20 '15 at 09:58
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$$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdot ... \cdot \frac{n}{n}\leq \frac{1}{n}\cdot\frac{2}{n}\cdot1\cdot ... \cdot 1= \frac{2}{n^2}$$

N. S.
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Here is a completely elementary proof with an explicit upper bound of 4:

$n! = \prod_{i=1}^n i$, so, reversing the order of the terms, $n! = \prod_{i=1}^n (n+1-i)$.

Multiplying these, $n!^2 = \prod_{i=1}^n i(n+1-i)$. But

$\begin{align} i(n+1-i) = i(n+1) - i^2 &= \frac{(n+1)^2}{4} - \frac{(n+1)^2}{4} +i(n+1) - i^2\\ &= \frac{(n+1)^2}{4} - (i-\frac{n+1}{2})^2\\ &\le \frac{(n+1)^2}{4} \end{align} $

so

$\begin{align} n!^2 &\le \prod_{i=1}^n {(n+1)^2}{4}\\ &= \big(\frac{(n+1)^2}{4}\big)^n\\ &= \big(\frac{n+1}{2}\big)^{2n}\\ \end{align} $

or $n! \le \big(\frac{n+1}{2}\big)^{n}$, so $$\frac{n!}{n^n} \le \big(\frac{n+1}{2n}\big)^n = \big(\frac1{2}+\frac1{2n}\big)^n$$

Since $\frac1{2}+\frac1{2n} \le \frac{3}{4}$ for $n \ge 2$,

$\begin{align} \sum_{n=1}^{\infty} \frac{n!}{n^n} &\le 1 + \sum_{n=2}^{\infty} \big(\frac{3}{4}\big)^n\\ &=1 + \frac{(3/4)^2}{1-3/4}\\ &= 1 + 9/4\\ &= 13/4\\ &< 4\\ \end{align} $.

marty cohen
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    I think this is an interesting answer but you should use \frac{a}{b} (between dollar signs, of course) to express a fraction instead of a/b, and also use double line space and double dollar sign to center and make things bigger and clear, for example compare: $\sum_{n=1}^\infty n!/n^n,$ with $$\sum_{n=1}^\infty\frac{n!}{n^n}$$ The first one is with one sign dollar to both sides, the ends one with &two& signs dollars at each side...As it is chances are that many people just don't took the time to try to read your answer because it looks messy. – DonAntonio Apr 09 '13 at 10:14
  • I know about those commands, but just was too lazy. However, when I did the reformatting as you suggested, I got two upvotes! Thanks. II will keep this in mind in the future. – marty cohen Apr 15 '13 at 06:05
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All you need is that $n! \approx c \sqrt{n}(n/e)^n$ for some real $c$.

Then $n!/n^n \approx c \sqrt{n}/e^n$ so the sum converges.

Actually, this shows that $\sum_{n=0}^{\infty} x^n n!/n^n$ converges for $|x| < e$.

marty cohen
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Use d'Alembert's ratio test, it's not hard to find the convergence of the series. http://en.wikipedia.org/wiki/Ratio_test

Phil Wang
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