# Thread: Help testing these series with any root test or ratio test.

1. ## Help testing these series with any root test or ratio test.

Problem 1 Evaluate

Σ ( 100^n (2n)! / (3n)! )
n=1

Converges or diverges?

And,

Problem 2 Evaluate

Σ [(n^3 + n)/(2n^3 + 5n^2 − 3n + 1)]^(n+1)
n=1

Converges or diverges?

2. nm

3. Originally Posted by Zocken
1.$\displaystyle \frac{100^2}{(n+1)!}*\frac{n!}{100^2}$

$\displaystyle =\frac{n!}{(n+1)!}=\frac{1}{n+1}\rightarrow 0$

so converges. might want to check me on the simplification of the permutations, not exactly sure on that.
I get the following using the ratio test:

$\displaystyle \lim_{n \to \infty} \Big| \frac {100^{n+1}(2n+2)!}{(3n+3)!} \frac {(3n)!}{(2n)!100^{n}} \Big|$ $\displaystyle = \lim_{n \to \infty} \frac {100(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)} = 0$ since the the numerator is a quadratic and the denominator is a cubic

4. the second one
use root test

$\displaystyle \left(\left(\frac{n^3+n}{2n^3+5n^2-3n+1}\right)^{n+1}\right)^{\frac{1}{n+1}}$

$\displaystyle lim_{n\rightarrow \infty}\frac{n^3+n}{2n^3+5n^2-3n+1}=\frac{1}{2}<1...converge$

5. nm

6. Hey thanks all for your help.

Random Variable:
There is a part that i dont understand which is when you went from

What happened with the (3n)! and the (2n)!?

7. Originally Posted by tapiaghector
Hey thanks all for your help.

Random Variable:
There is a part that i dont understand which is when you went from

What happened with the (3n)! and the (2n)!?
$\displaystyle lim_{n\rightarrow\infty}\left(\frac{100^{n+1}(2n+2 )!}{(3n+3)!}\frac{(3n)!}{(2n)!100^{n}}\right)$

$\displaystyle lim_{n\rightarrow\infty}\left(\frac{100^{n+1}(2n+2 )(2n+1)(2n)!}{(3n+3)(3n+2)(3n+1)(3n)!}\frac{(3n)!} {(2n)!100^{n}}\right)$

as you can see (2n)! divide (2n)! =1 the same thing to (3n)!

8. Originally Posted by tapiaghector
Hey thanks all for your help.

Random Variable:
There is a part that i dont understand which is when you went from

What happened with the (3n)! and the (2n)!?
$\displaystyle \frac {100^{n+1} (2n+2)(2n+1)(2n)!}{(3n+3)(3n+2)(3n+1)(3n)!} \frac {(3n)!}{(2n!) 100^{n}}$