$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}$

determine whether the series converges or diverges

help?(Speechless)

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- Mar 21st 2009, 07:29 PMbuttonbearseries with factorial
$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}$

determine whether the series converges or diverges

help?(Speechless) - Mar 22nd 2009, 05:34 AMSoroban
Hello, buttonbear!

Without the Ratio Test, I'm not sure of the proof.

Quote:

$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}$ . Determine whether the series converges or diverges

$\displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}$

. . $\displaystyle = \;\frac{n+1}{1}\cdot\frac{n^n}{(n+1)^{n+1}} \;=\;\frac{n^n}{(n+1)^n} \;=\;\left(\frac{n}{n+1}\right)^n$

Divide top and bottom of the fraction by $\displaystyle n\!:\;\;\left(\frac{1}{1+\frac{1}{n}}\right)^n \;=\;\frac{1}{\left(1 + \frac{1}{n}\right)^n} $

Therefore: .$\displaystyle \lim_{n\to\infty}R \;=\;\lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{\lim\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{e} \;<\;1$

. . and the series converges.

- Mar 22nd 2009, 09:22 AMo_O
For all $\displaystyle n \geq 2$, we have: $\displaystyle \frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots (n-1)n}{n \cdot n \cdot n \cdots n \cdot n} \ \ {\color{red} \leq}\ \ \frac{1}{n} \cdot \frac{2}{n} \cdot 1 \cdot 1 \cdots \cdot 1 = \frac{2}{n^2}$

So by comparison ... - Mar 22nd 2009, 03:35 PMbuttonbear
i'm sorry, i just didn't fully understand your post..

- Mar 22nd 2009, 05:23 PMo_O
I've showed that: $\displaystyle \frac{n!}{n^n} < \frac{2}{n^2}$

Since $\displaystyle \sum \frac{2}{n^2}$ is convergent, by the comparison test ... what can you conclude? - Mar 22nd 2009, 05:44 PMbuttonbear
i can conclude that... because the larger sequence converges..the other one does too?

- Mar 22nd 2009, 06:04 PMbuttonbear
i'm just kind of confused as how to how you knew to pick $\displaystyle \frac{2}{n^2}$ for your larger sequence? or can it be any larger sequence?

- Aug 13th 2009, 02:28 PMjohntuan
sorry for bumping up an old thread but how did you know to compare it to 2/n^2 as opposed to 1/n^2??

- Aug 13th 2009, 02:39 PMjohntuan
oh nvm i think i see it, because you cant use 1/n^2, you can only use 2/n^2, cause if you only use the first term it would be 1/n, which is divergent according to the p-series, and so you use n>=2..ahh i see it..