2

I know that this series diverges.

$$\sum_{n=1}^\infty (-1)^n \frac{n^n}{n!}$$

From what I learned, as it's an alternating series, I have to prove that $\frac{n^n}{n!}$ descends and $\lim_{n\to \infty} \frac{n^n}{n!} = 0$.

However, I don't know what method I should use to break down $\frac{n^n}{n!}$.

Thanks for your help.

(Also it's my first time editing formally like this, so if there's any mistake, any correction is appreciated!)

Blue
  • 75,673
Pearly
  • 47

4 Answers4

3

Since $n^n>n!$ as easily seen by induction you have that if you call with $a_n$ the general term of your series, it does not have limit $0$ as $n \to \infty$.Therefore the series is divergent.

Luca Goldoni Ph.D.
  • 2,969
1

HINT. Apply the Ratio Test carefully.

As a general note, some series which are alternating, are much easier shown to be convergent using the Root/Ratio Test (at least those which are absolutely convergent).

mathematics2x2life
  • 13,322
  • After trying it for yourself, check out: https://math.stackexchange.com/questions/354487/does-the-series-sum-n-nn-converge-or-diverge/354491 and https://socratic.org/questions/how-do-you-apply-the-ratio-test-to-determine-if-sigma-n-n-n-from-n-1-oo-is-conve Your series is handled no differently as if $\sum |a_n|$ converges, so too does $\sum a_n$. – mathematics2x2life Feb 17 '20 at 02:58
  • https://socratic.org/questions/how-do-you-apply-the-ratio-test-to-determine-if-sigma-n-n-n-from-n-1-oo-is-conve – mathematics2x2life Feb 17 '20 at 02:59
1

Stirling's formula for $n!$ is $$n! \approx \sqrt{2\pi} \, n^{n+1/2} e^{-n}$$

So the $n^\textrm{th}$ term of your series is approximately

$$a_n \approx (-1)^n \frac{e^n}{\sqrt{2\pi n} }.$$

The root test gives $|a_n|^{1/n} \rightarrow e >1.$

So the series diverges.

mjw
  • 8,647
  • 1
  • 8
  • 23
1

A necessary condition for any series $\sum_{n=1}^\infty a_n$ to converge is that $\lim_{n\to\infty} |a_n| =0$.

In your case is $a_n = (-1)^n \frac{n^n}{n!}$ and it is easy to verify that this sequence does not converge. For $n >1$ you have

$$|a_n| = \frac{\overbrace{n \cdots n}^{\lfloor \frac n2 \rfloor\; factors} \cdot n \cdots n}{1 \cdots \lfloor \frac n2 \rfloor \cdot (\lfloor \frac n2 \rfloor +1) \cdots n} > 2^{\lfloor \frac n2 \rfloor}\stackrel{n\to \infty}{\longrightarrow}+\infty$$ So, your series does not converge.

trancelocation
  • 32,243