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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.

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1. watchmath

Use ratio test.

2. 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.

3. anonymous

Thanks :D

4. 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?

5. watchmath

Yes, now $$\frac{n+1}{n}=1+\frac{1}{n}$$

6. anonymous

Would you mind giving a detail of how to factor that?

7. 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 $n\to\infty$

8. watchmath

Since $$n>0$$ the expression $$1+\frac{1}{n}$$ is positive. So we don't need absolute value.

9. anonymous

Then you apply L'hopital correct?

10. watchmath

yes you may. But actually that is how we usually define the natural number $$e$$.

11. anonymous

Ah OK, if you remember the formula. Awesome, thanks again.

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