## anonymous 5 years ago i really need the answer.. how will i solve for the series n!/n^n to show it is converging/diverging

1. anonymous

it converges

2. anonymous

use the ratio test

3. anonymous

$\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}$=$\frac{(n+1)n!}{(n+1)^n(n+1)}\frac{n^n}{n!}$=$\frac{n^n}{(n+1)^n}$=$\left(\frac{n}{n+1}\right)^n$=$(1-\frac{1}{n+1})^n$

4. anonymous

now $\lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right)^n$

5. anonymous

i did this. we get 1 by this and it means that the test fails

6. anonymous

no you get $frac{1}{e}$

7. anonymous

$\frac{1}{e}$

8. anonymous

if you try and take the limit directly you get the indeterminate form $1^\infty$ not 1

9. anonymous

you can evaluate the limit by capitalizing on the continuity of the natural logarithm function

10. anonymous

like so first set $(1-\frac{1}{n+1})^n=L$ apply the natural log function to both sides to get $ln (1-\frac{1}{n+1})^n=\ln L$

11. anonymous

sorry I should be writing limit on the LHS

12. anonymous

even then the answer will be 1

13. anonymous

no its not

14. anonymous

how will you solve it?

15. anonymous

so we now use properties of logarithms

16. anonymous

$\lim_{n\rightarrow\infty}n\ln(1-\frac{1}{n+1})=\ln L$ which still yields yet another indeterminate form $\infty \times 0$ now rewrite to the equivalent expression $\lim_{n\rightarrow\infty}\frac{\ln(1-\frac{1}{n+1})}{\frac{1}{n}}=\ln L$ so now we have indeterminate form $\frac{0}{0}$ and now can finally use L'Hopital's rule

17. anonymous

apply L'Hopital's rule once we get $\lim_{n\rightarrow\infty}\frac{\frac{1}{1-\frac{1}{n+1}}\frac{1}{(n+1)^2}}{-\frac{1}{n^2}}=\ln L$ simplifying we get $\lim_{n\rightarrow\infty}\frac{-n^2}{(n+1)^2-(n+1)}=\ln L$ finally taking the limit we see that $-1=\ln L$

18. anonymous

exponentiating we see $L=\frac{1}{e}$

19. anonymous

that is a common result and easily verified with software

20. anonymous

ok.. thanks alot