# Math Help - Convergence / Divergence

1. ## Convergence / Divergence

series converge or diverge?

Σ ((−1)^i*(i)^2)/(i^4)*-1
i=2

I have tryed using Alternating Series Test because when the sums are worked out eg i=2 (+) number where i-3 (-) number.

- But cant figure out where they Convergence OR Divergence.
- Is there another test i can use?

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Σ (n^2)/(2n +1)!
n=0

Can the ratio test be used?

I am quite confused....

2. Originally Posted by wrx12
series converge or diverge?

Σ ((−1)^i*(i)^2)/(i^4)*-1
i=2

I have tryed using Alternating Series Test because when the sums are worked out eg i=2 (+) number where i-3 (-) number.

- But cant figure out where they Convergence OR Divergence.
- Is there another test i can use?

----------------------------------------------------------------------------

Σ (n^2)/(2n +1)!
n=0

Can the ratio test be used?

I am quite confused....

if the first one is this

$\sum_{i=2}^{\infty}\frac{(-1 )^i(i^2)}{i^4 (-1)}=\sum_{i=2}^{\infty}\frac{(-1)^{i-1}}{i^2}$

This converges by the AST.

For the second the ratio test will work.

Good luck.

3. ## Test

Thanks will try and see what i get..

4. ## For the second one

you could also use the integral test(I'm saying this mainly because I like the integral test..even though in this case its hard =) so you need to determine if $\int_1^{\infty}\frac{x^2}{2x+1}$ exists...you can evaluate this integral using partial fraction decomposition...so $\int_1^{\infty}\frac{x^2}{2x+1}=\int_1^{\infty}\fr ac{1}{8x+4}+\frac{x}{2}-\frac{1}{4}=\bigg[\frac{\ln(2x+1)}{8}+\frac{x^2}{4}-\frac{x}{4}\bigg]|_1^{\infty}$..which is obviously $\infty$..therfore since the integral diverges the series diverges...I am sorry I was bored

5. Hello mathstud28

For your integral, it's far more simple to say that at infinite, the function inside the integral tends to infinity. Thus the integral is undefined.
The sole problem is that it was (2n+1)!, not only (2n+1).

This method is too complicated while a ratio test can give the answer. Although it's elegant, it's not judicious to use it.

Ratio test :

$a_n=\frac{n^2}{(2n+1)!}$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2 (2n+1)!}{n^2 (2n+3)!}=\frac{(n+1)^2}{n^2 (2n+2)(2n+3)}$ $=\frac{n+1}{2n^2 (2n+3)}$

Which is always < 1, so the series converges

6. ## Yeah

that is why I had a disclaimer in the beginning about the integral test being overly difficult here...I just wanted to show that never forget the lesser used tests...and that most can be done by more than one test if not most of them

7. Originally Posted by Mathstud28
that is why I had a disclaimer in the beginning about the integral test being overly difficult here...I just wanted to show that never forget the lesser used tests...and that most can be done by more than one test if not most of them
You quoted the integral test for doing a wrong thing, i haven't seen any disclaimer oO
I'm not sure to understand, what d'you want to say in this message ?
The lesser used tests can give some bad surprises, especially when you're not used to apply it, more especially when there is a test which is more simple !

8. ## O

you are saying I missed the fact that it was factorial...yeah you are right I do that a lot when its not LaTeX its hard to discern...nonetheless I was trying to give dimension to a problem that is generally one-dimensional...thanks for pointing out I missed the ! though!

9. I think i didn't understand well what happened, forget it