# Thread: Series problem

1. ## n =0

I Hope this might help u to read better,

assess the series

infinity

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

n = 0

2. Don't post the same question multipe times

RonL

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

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

5. $\displaystyle \sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$
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.

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

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.

7. ## sorry !

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

8. 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.

9. ## thanks a lot

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

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

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.

11. 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 15th Post!!!

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

If the limit is one, this most certainly diverges.

13. 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.

14. 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.