# Thread: Integral Test and comparison test help!

1. ## Integral Test and comparison test help!

Hello, I have been looking around for how to solve these to series by the integral test.

Problem 1 Evaluate

Σ (n^(1/n))/n^3
n=1

decide if the series converges or diverges?

And,

Problem 2 Evaluate

Σ (3^n*n!)/(3k)!
n=1

decide if the series converges or diverges?

2. $\sum_{n=1}^{\infty}\frac{n^{\frac{1}{n}}}{n^3}$

$\sum_{n=1}^{\infty}\frac{3^n(n!)}{(3n)!}$

the second one by root test

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

$lim_{n\rightarrow\infty}\frac{3(n!)^{\frac{1}{n}}} {((3n)!)^{\frac{1}{n}}}=3$

limit is 3>1 so the series diverge the first one sorry I do not know how to decide yet ..

3. Hey thanks for the help but in the denominator there's a k instead of an n. That's the trick question!

4. k is constant or what ???
I am sorry
it is higher level

5. Originally Posted by tapiaghector
Hello, I have been looking around for how to solve these to series by the integral test.

Problem 1 Evaluate

Σ (n^(1/n))/n^3
n=1

decide if the series converges or diverges?

And,

Problem 2 Evaluate

Σ (3^n*n!)/(3k)!
n=1

decide if the series converges or diverges?
First question - why on earth would you want to use the integral test (remember, you will need to integrate).

I would personally use a limit comparison test for the first. Compare with

$
\sum_{n=1}^{\infty} \frac{1}{n^3}
$

for the second (it looks a little fishy) but

$\frac{1}{(3k)!} \sum_{n=1}^{\infty} 3^n n!$ which clearly diverges - the terms don't go to zero.

6. So, for the second one: how are you sure that the 3^n*n! diverges since the you dont know 1/3k!. Also, if you plug numbers in the series it goes to larger and larger...

there isnt another method to compared this serie?

7. Originally Posted by tapiaghector
So, for the second one: how are you sure that the 3^n*n! diverges since the you dont know 1/3k!. Also, if you plug numbers in the series it goes to larger and larger...

there isnt another method to compared this serie?
k does not depend on n and so is treated as a constant. So when dealing with the summation you only have to worry about the stuff that depends on n.