Question

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

Answer 1

it converges

Answer 2

use the ratio test

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

no you get \[frac{1}{e}\]

Answer 7

\[\frac{1}{e}\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

sorry I should be writing limit on the LHS

Answer 12

even then the answer will be 1

Answer 13

no its not

Answer 14

how will you solve it?

Answer 15

so we now use properties of logarithms

Answer 16

\[\lim_{n\rightarrow\infty}n\ln(1-\frac{1}{n+1})=\ln L\] 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

Answer 17

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 \[-1=\ln L\]

Answer 18

exponentiating we see \[L=\frac{1}{e}\]

Answer 19

that is a common result and easily verified with software

Answer 20

ok.. thanks alot