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

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

Mathematics
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it converges
use the ratio test
\[\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}\]=\[\frac{(n+1)n!}{(n+1)^n(n+1)}\frac{n^n}{n!}\]=\[\frac{n^n}{(n+1)^n}\]=\[\left(\frac{n}{n+1}\right)^n\]=\[(1-\frac{1}{n+1})^n\]

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Other answers:

now \[\lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right)^n\]
i did this. we get 1 by this and it means that the test fails
no you get \[frac{1}{e}\]
\[\frac{1}{e}\]
if you try and take the limit directly you get the indeterminate form \[1^\infty\] not 1
you can evaluate the limit by capitalizing on the continuity of the natural logarithm function
like so first set \[(1-\frac{1}{n+1})^n=L\] apply the natural log function to both sides to get \[ln (1-\frac{1}{n+1})^n=\ln L\]
sorry I should be writing limit on the LHS
even then the answer will be 1
no its not
how will you solve it?
so we now use properties of logarithms
\[\lim_{n\rightarrow\infty}n\ln(1-\frac{1}{n+1})=\ln L\] 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
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 \[-1=\ln L\]
exponentiating we see \[L=\frac{1}{e}\]
that is a common result and easily verified with software
ok.. thanks alot

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