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Convergence

Alexmahone

Active member
Jan 26, 2012
268
Do the following series converge or diverge?

a) $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n$

b) $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$

My attempt:

a) $\displaystyle\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx\frac{1}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{1}{\sqrt{2\pi n}}$

Since $\displaystyle\sum\frac{1}{\sqrt{2\pi n}}$ diverges, $\displaystyle \sum\frac{1}{n!}\left(\frac{n}{e}\right)$ also diverges.

b) $\displaystyle\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx\frac{(-1)^n}{\sqrt{2\pi n}}$ (Stirling’s formula)

$\displaystyle \sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n\approx \sum\frac{(-1)^n}{\sqrt{2\pi n}}$

By Leibniz's test for alternating series, $\displaystyle\sum\frac{(-1)^n}{\sqrt{2\pi n}}$ converges.

So, $\displaystyle\sum\frac{(-1)^n}{n!}\left(\frac{n}{e}\right)^n$ also converges.
 
Last edited:

Alexmahone

Active member
Jan 26, 2012
268
Is my solution correct?
 

Alexmahone

Active member
Jan 26, 2012
268
Setting $\displaystyle a_{n}= \frac{1}{n!}\ (\frac{n}{e})^{n}$ You have...

$\displaystyle \ln a_{n}= n\ (\ln n -\ln e) -\ln n!= n\ (\ln n-1) - \sum_{k=1}^{n} \ln k$ (1)

Now You have...

$\displaystyle n\ (\ln n-1)= \int_{1}^{n} \ln x\ dx > \sum_{k=1}^{n} \ln k$ (2)

... so that is...

$\displaystyle \lim_{n \rightarrow \infty} \ln a_{n} >0 \implies \lim_{n \rightarrow \infty} a_{n} >1$ (3)

... and the necessary condition for convergence is not verified...

Kind regards

$\chi$ $\sigma$
But what's wrong with my solution?
 

chisigma

Well-known member
Feb 13, 2012
1,704
My previous post was wrong because I incorrectly wrote $\displaystyle \int_{1}^{n} \ln x\ dx > \sum_{k=1}^{n} \ln k$ and is $\displaystyle \int_{1}^{n} \ln x\ dx < \sum_{k=1}^{n} \ln k$...

Very sorry!... any thanks is absolutely a nonsense!...

Kind regards

$\chi$ $\sigma$
 
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Alexmahone

Active member
Jan 26, 2012
268
Writing $\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt{2\ \pi\ n}}{n!}\ (\frac{n}{e})^{n}=1$ doesn't mean $\displaystyle \frac{1}{n!}\ (\frac{n}{e})^{n}= \frac{1}{\sqrt{2\ \pi\ n}}$...

Kind regards

$\chi$ $\sigma$
But aren't they approximately equal for large $n$?

---------- Post added at 03:27 AM ---------- Previous post was at 03:09 AM ----------

My previous post was wrong because I incorrectly wrote $\displaystyle \int_{1}^{n} \ln x\ dx > \sum_{k=1}^{n} \ln k$ and is $\displaystyle \int_{1}^{n} \ln x\ dx < \sum_{k=1}^{n} \ln k$...

Very sorry!... any thanks is absolutely a nonsense!...

Kind regards

$\chi$ $\sigma$
Does that mean my solution is right?
 

chisigma

Well-known member
Feb 13, 2012
1,704
But aren't they approximately equal for large $n$?

---------- Post added at 03:27 AM ---------- Previous post was at 03:09 AM ----------



Does that mean my solution is right?
Rewriting my previous post we have...

$\displaystyle \int_{1}^{n} \ln x\ dx - \sum_{k=1}^{n} \ln k = \sum_{k=1}^{n} \int_{k}^{k+1} (ln x - \ln (k+1))\ dx \sim - \sum_{k=1}^{n} \frac{1}{k+1}$

... so that is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and Your solution is probably correct...

Kind regards

$\chi$ $\sigma$