I Hope this might help u to read better,

assess the series

infinity

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

n = 0

Printable View

- Jun 22nd 2006, 10:29 AMdistancen =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 PMCaptainBlack
Don't post the same question multipe times

RonL - Jun 22nd 2006, 05:09 PMdistance
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 PMJameson
Is your second thread titled "More series problems" something you would like deleted?

- Jun 22nd 2006, 06:23 PMThePerfectHacker
$\displaystyle \sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$

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

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 AMdistancesorry !
$\displaystyle

\sum_{n=0}^{\infty}\frac{(-2)^n}{(3n+1)!}

$ - Jun 23rd 2006, 08:34 AMJameson
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 AMdistancethanks a lotQuote:

Originally Posted by**Jameson**

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

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

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 AMThePerfectHackerQuote:

Originally Posted by**Jameson**

This is my 15:):)th Post!!! - Jun 23rd 2006, 11:10 AMJameson
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 PMThePerfectHacker
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 PMJameson
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.