Series problem

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June 22nd 2006, 10:29 AM
distance

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n =0

I Hope this might help u to read better,

assess the series

infinity

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

n = 0
June 22nd 2006, 01:11 PM
CaptainBlack

Don't post the same question multipe times

RonL
June 22nd 2006, 05:09 PM
distance

Sorry about that. Anyone can do this problem ( bouncing ball), the final is in next week, so i need to understand the problem
June 22nd 2006, 05:41 PM
Jameson

Is your second thread titled "More series problems" something you would like deleted?
June 22nd 2006, 06:23 PM
ThePerfectHacker

$\sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$
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.
June 23rd 2006, 03:50 AM
Jameson

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

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.
June 23rd 2006, 06:01 AM
distance

sorry !

$<br /> \sum_{n=0}^{\infty}\frac{(-2)^n}{(3n+1)!}<br />$
June 23rd 2006, 08:34 AM
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.
June 23rd 2006, 08:44 AM
distance

thanks a lot

Quote:

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.

$<br /> \sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}<br />$
June 23rd 2006, 09:01 AM
Jameson

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

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.
June 23rd 2006, 09:06 AM
ThePerfectHacker

Quote:

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!!!
June 23rd 2006, 11:10 AM
Jameson

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

If the limit is one, this most certainly diverges.
June 23rd 2006, 12:05 PM
ThePerfectHacker

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.
June 23rd 2006, 01:28 PM
Jameson

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.