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anonymous
 5 years ago
Determine if \[\sum_{n=1}^{\infty} (1)^n \left( \frac{n^n}{n!} \right)\] diverges, converges conditionally or converges absolutely.
anonymous
 5 years ago
Determine if \[\sum_{n=1}^{\infty} (1)^n \left( \frac{n^n}{n!} \right)\] diverges, converges conditionally or converges absolutely.

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(a_{n+1}/a_n=\frac{(n+1)^{n+1}n!}{(n+1)!n^n}=(\frac{n+1}{n})^n=(1+\frac{1}{n})^n\) The limit is \(e\). So the series diverges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So this is what I did: \[\lim_{n\to\infty}\left\frac{\frac{(1)^{n+1}(n+1)^{n+1}}{(n+1)!}}{\frac{(1)^{n}n^{n}}{n!}}\right\]\[\lim_{n\to\infty}\left\frac{(1)^{n+1}(n+1)^{n+1}n!}{(1)^{n}n^{n}(n+1)!}\right\]\[\lim_{n\to\infty}\left\frac{(n+1)^{n}}{n^{n}}\right\]\[\lim_{n\to\infty}\left\left(\frac{n+1}{n}\right)^{n}\right\] How do I continue?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, now \(\frac{n+1}{n}=1+\frac{1}{n}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would you mind giving a detail of how to factor that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, it should've been obvious :D \[\frac{n}{n}+\frac{1}{n}\] So you remove the absolute value just for the fact that \[n\to\infty\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Since \(n>0\) the expression \(1+\frac{1}{n}\) is positive. So we don't need absolute value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you apply L'hopital correct?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1yes you may. But actually that is how we usually define the natural number \(e\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah OK, if you remember the formula. Awesome, thanks again.
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