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?

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- Jun 4th 2009, 07:24 PMtapiaghectorHelp 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? - Jun 4th 2009, 07:50 PMZocken
nm

- Jun 4th 2009, 08:08 PMRandom Variable
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 - Jun 4th 2009, 08:17 PMAmer
the second one

use root test

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

$\displaystyle lim_{n\rightarrow \infty}\frac{n^3+n}{2n^3+5n^2-3n+1}=\frac{1}{2}<1...converge$ - Jun 4th 2009, 08:18 PMZocken
nm

- Jun 4th 2009, 09:19 PMtapiaghector
Hey thanks all for your help.

Random Variable:

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

http://www.mathhelpforum.com/math-he...a03890f5-1.gif http://www.mathhelpforum.com/math-he...90e29080-1.gif

What happened with the (3n)! and the (2n)!? - Jun 4th 2009, 09:31 PMAmer
$\displaystyle lim_{n\rightarrow\infty}\left(\frac{100^{n+1}(2n+2 )!}{(3n+3)!}\frac{(3n)!}{(2n)!100^{n}}\right)$

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

as you can see (2n)! divide (2n)! =1 the same thing to (3n)! - Jun 4th 2009, 09:33 PMRandom Variable