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

**tapiaghector** · June 4th 2009, 07:24 PM

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?

**Zocken** · June 4th 2009, 07:50 PM

nm

**Random Variable** · June 4th 2009, 08:08 PM

Originally Posted by Zocken

1. $\frac{100^2}{(n+1)!}*\frac{n!}{100^2}$

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

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:

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

**Amer** · June 4th 2009, 08:17 PM

the second one
use root test

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

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

**Zocken** · June 4th 2009, 08:18 PM

nm

**tapiaghector** · June 4th 2009, 09:19 PM

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)!?

**Amer** · June 4th 2009, 09:31 PM

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)!?

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

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

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

**Random Variable** · June 4th 2009, 09:33 PM

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)!?

$\frac {100^{n+1} (2n+2)(2n+1)(2n)!}{(3n+3)(3n+2)(3n+1)(3n)!} \frac {(3n)!}{(2n!) 100^{n}}$

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