anonymous
  • anonymous
Determine if \[\sum_{n=1}^{\infty} (-1)^n \left( \frac{n^n}{n!} \right)\] diverges, converges conditionally or converges absolutely.
Mathematics
schrodinger
  • schrodinger
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watchmath
  • watchmath
Use ratio test.
watchmath
  • watchmath
\(|a_{n+1}/a_n|=\frac{(n+1)^{n+1}n!}{(n+1)!n^n}=(\frac{n+1}{n})^n=(1+\frac{1}{n})^n\) The limit is \(e\). So the series diverges.
anonymous
  • anonymous
Thanks :D

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anonymous
  • anonymous
So 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
  • watchmath
Yes, now \(\frac{n+1}{n}=1+\frac{1}{n}\)
anonymous
  • anonymous
Would you mind giving a detail of how to factor that?
anonymous
  • anonymous
Oh, 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
  • watchmath
Since \(n>0\) the expression \(1+\frac{1}{n}\) is positive. So we don't need absolute value.
anonymous
  • anonymous
Then you apply L'hopital correct?
watchmath
  • watchmath
yes you may. But actually that is how we usually define the natural number \(e\).
anonymous
  • anonymous
Ah OK, if you remember the formula. Awesome, thanks again.

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