# Series problem

• Jun 22nd 2006, 10:29 AM
distance
n =0
I Hope this might help u to read better,

assess the series

infinity

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

n = 0
• Jun 22nd 2006, 01:11 PM
CaptainBlack
Don't post the same question multipe times

RonL
• Jun 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
• Jun 22nd 2006, 05:41 PM
Jameson
Is your second thread titled "More series problems" something you would like deleted?
• Jun 22nd 2006, 06:23 PM
ThePerfectHacker
$\displaystyle \sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$
Use the generalized ratio test,
$\displaystyle \lim_{n\to\infty}\left| \frac{(-2)^{n+1}}{(3(n+1)+1)!}\cdot \frac{(3n+1)!}{(-2)^n}\right|$
Note that,
$\displaystyle (3(n+1)+1)!=(3n+4)!=(3n+4)(3n+3)(3n+2)(3n+1)!$
Thus,
$\displaystyle \left| \frac{-2}{(3n+4)(3n+3)(3n+2)} \right|= 0$ as $\displaystyle n\to\infty$
Which is less than 1 thus the series converges.
---
Also, the sequence,
$\displaystyle \left\{ \frac{(-2)^n}{(3n+1)!} \right\}$
Fits the conditions for Leibniz's alternating series test, thus it absolutely convergent.
• Jun 23rd 2006, 03:50 AM
Jameson
$\displaystyle \sum_{n=0}^{\infty}\frac{n!}{e^n}$

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

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

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

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

Thus this diverges.

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

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

If this limit is anything but zero, the series diverges. If it is zero, you must use another test to help you.
• Jun 23rd 2006, 06:01 AM
distance
sorry !
$\displaystyle \sum_{n=0}^{\infty}\frac{(-2)^n}{(3n+1)!}$
• Jun 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.
• Jun 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.

$\displaystyle \sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}$
• Jun 23rd 2006, 09:01 AM
Jameson
$\displaystyle \sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}$

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

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

Thus this diverges.
• Jun 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!!!
• Jun 23rd 2006, 11:10 AM
Jameson
What about this series then... $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ :D :D

If the limit is one, this most certainly diverges.
• Jun 23rd 2006, 12:05 PM
ThePerfectHacker
I do not understand you?
The ratio test is inconlusive for $\displaystyle L=1$.
In your other post you said for $\displaystyle L=0$,
I was making a correction.
• Jun 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.