Analysis of a_n Series: Convergence or Divergence?

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azatkgz
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Suppose that [tex]a_n\geq 0[/tex] and there is

[tex]\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=c[/tex]
If c>1,series diverges.
if c<1 series converges.

For [tex]a_n=\frac{n!}{n^n}[/tex]

[tex]\lim_{n\rightarrow\infty}\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}[/tex]

[tex]\lim_{n\rightarrow\infty}\frac{n^n}{(n+1)^n}[/tex]

Then I used I'Hopital Rule and got answer 1.
 
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I think you'd better check your l'Hopital. Your final limit is closely related to the limit of (1+1/n)^n, which is a famous limit and is not one.
 
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A,yes I get it.


[tex]\lim_{n\rightarrow\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}[/tex]