Question

Does this series converge?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ n! }{ n^n }\]

Answer 2

series:
Σ n! / (n^n) ... n = 1 to ∞

use ratio test to determine convergence:

lim |Un+1 / Un|
= lim [(n+1)! / (n+1)^(n+1)] / [n! / n^n]
= lim (n+1) n^n / (n+1)^(n+1)
= lim n^n / (n+1)^n
= lim (n/(n+1))^n
= lim e^ln (n/(n+1))^n
= e ^ lim ln (n/(n+1))^n
= e ^ lim n * ln (n/(n+1))
= e ^ lim [ln(1 - 1/(n+1))] / [1/n]

and using L'Hopital's rule
= e ^ lim [1/(1 - 1/(n+1)) * (1/(n+1)^2] / [-1/n^2]
= e ^ - lim 1/(1 - 1/(n+1)) * lim (n/(n+1))^2
= e ^ (-1) * 1
= 1/e

and this ratio is less than 1.
thus the series is convergent .

Answer 3

This is how i solved it.

Answer 4

...first let me get some icecream.

Answer 5

D:
I want ice cream..
And thank you @FutureMathProfessor

Answer 6

Any time Luigi. I'll be your mario any day.

Answer 7

lol?

Answer 8

\(\sum\limits_{n=1}^\infty\frac{n!}{n^n}\)
Let's use the ratio test.$$\begin{align*}\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^n}{n!}&=\lim_{n\to\infty}\frac{(n+1)n^n}{(n+1)^{n+1}}\\&=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}\\&=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\\&=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)^n\\&=\lim_{n\to\infty}\left(1-\frac1n\right)^n\\&=\frac1e<1\end{align*}$$

Answer 9

\[\large \sum_{n=1}^{\infty}\frac{n!}{n^n}\] for factorials and exponents, I would use a ratio test to find whether the series converges or diverges.

Ratio Test: \[\large \lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}| = L < 1 \ \therefore \ \sum_{n=1}^{\infty}a_{n} = C\]
Define your \(\large a_{n+1}\) and your \(\large a_{n}\)
\[\large a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}\]\[\large a_{n} = \frac{n!}{n^n}\]now take the limit of these two. \[\large \lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_{n}}|\]\[\large \lim_{n \rightarrow \infty}|\frac{(n+1)!}{(n+1)^{n+1}} * \frac{n^{n}}{(n)!}|\]expand it out \[\large \lim_{n \rightarrow \infty}|\frac{(n+1)n!}{(n+1)*(n+1)^{n}}*\frac{n^n}{n!}|\] cancel out like terms\[\large \lim_{n \rightarrow \infty}|\frac{n^n}{(n+1)^n}|\] put the fraction under one power. \[\large \lim_{n \rightarrow \infty}|(\frac{n}{n+1})^{n}|\]Now if you try taking the limit as n goes to infinity, you'll get \[\large \lim_{n \rightarrow \infty}|(\frac{n}{n+1})^{n}| = (\frac{\infty}{\infty + 1})^{\infty}= \infty\] So we can take the reciprocal of the fraction and put it under 1, because we know if we get infinity in the denominator, 1/infinity = 0. \[\large \lim_{n \rightarrow \infty}|\frac{1}{(\frac{n+1}{n})^{n}}|\] we also know that \[\large (\frac{n+1}{n})^{n} = (1+\frac1n)^n = e\] so \[\large \lim_{n \rightarrow \infty}|\frac{1}{(\frac{n+1}{n})^{n}}| = \frac{1}{e}\]\[\large \text{by definition}: \large \lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}| = \frac{1}{e} < 1 \ \therefore \ \sum_{n=1}^{\infty}a_{n} = C\]

Answer 10

beautiful :D

Answer 11

I wouldn't expect anything less from Miss Jhann