Series problem

**distance** · June 22nd 2006, 10:29 AM

I Hope this might help u to read better,

assess the series

infinity

sum of ((-2)^n)/(3n+1)!

n = 0

**CaptainBlack** · June 22nd 2006, 01:11 PM

Don't post the same question multipe times

RonL

**distance** · June 22nd 2006, 05:09 PM

Sorry about that. Anyone can do this problem ( bouncing ball), the final is in next week, so i need to understand the problem

**Jameson** · June 22nd 2006, 05:41 PM

Is your second thread titled "More series problems" something you would like deleted?

**ThePerfectHacker** · June 22nd 2006, 06:23 PM

$\sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$
Use the generalized ratio test,
$\lim_{n\to\infty}\left| \frac{(-2)^{n+1}}{(3(n+1)+1)!}\cdot \frac{(3n+1)!}{(-2)^n}\right|$
Note that,
$(3(n+1)+1)!=(3n+4)!=(3n+4)(3n+3)(3n+2)(3n+1)!$
Thus,
$\left| \frac{-2}{(3n+4)(3n+3)(3n+2)} \right|= 0$ as $n\to\infty$
Which is less than 1 thus the series converges.
---
Also, the sequence,
$\left\{ \frac{(-2)^n}{(3n+1)!} \right\}$
Fits the conditions for Leibniz's alternating series test, thus it absolutely convergent.

**Jameson** · June 23rd 2006, 03:50 AM

$\sum_{n=0}^{\infty}\frac{n!}{e^n}$

Just analytically thinking about this, I can see that the factorial will very strongly outgrow the exponential as x gets larger. But more formally...

$R_{n}=\frac{(n+1)!}{e^n \times e} \times \frac{e^n}{n!}$

$R_{n}=\frac{n}{e}$

$\lim_{n \to \infty}R_{n}=\infty$

Thus this diverges.

Another way is to just take the limit of the original series.

$\lim_{n \to \infty} \frac{n!}{e^n}=\infty$

If this limit is anything but zero, the series diverges. If it is zero, you must use another test to help you.

**distance** · June 23rd 2006, 06:01 AM

$<br /> \sum_{n=0}^{\infty}\frac{(-2)^n}{(3n+1)!}<br />$

**Jameson** · June 23rd 2006, 08:34 AM

ThePerfectHacker showed you how to show this converges. What's the problem? Don't worry about the lower bound of one versus zero. His work is still correct.

**distance** · June 23rd 2006, 08:44 AM

Originally Posted by Jameson

ThePerfectHacker showed you how to show this converges. What's the problem? Don't worry about the lower bound of one versus zero. His work is still correct.

$<br /> \sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}<br />$

**Jameson** · June 23rd 2006, 09:01 AM

$\sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}$

Always start by taking the limit of the series. If it's non-zero then the series diverges.

$\lim_{n \to \infty} \frac{n-1}{(3n+1)} = \frac{1}{3}$

Thus this diverges.

**ThePerfectHacker** · June 23rd 2006, 09:06 AM

Originally Posted by Jameson

If it is zero, you must use another test to help you.

Correction, 'if it is one' then you should use a different test.
This is my 15

th Post!!!

**Jameson** · June 23rd 2006, 11:10 AM

What about this series then... $\sum_{n=1}^{\infty}\frac{1}{n}$

If the limit is one, this most certainly diverges.

**ThePerfectHacker** · June 23rd 2006, 12:05 PM

I do not understand you?
The ratio test is inconlusive for $L=1$ .
In your other post you said for $L=0$ ,
I was making a correction.

**Jameson** · June 23rd 2006, 01:28 PM

Oh! Ok. I was talking about the nth term test. If the nth term does not approach 0, then the series diverges. Yes you are correct about the Ratio Test.

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n =0

sorry !

thanks a lot

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