- #1
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.
[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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