Converge/Diverge: Summation 1 to ∞ (2*4*6*8*...*(2n)) / (n!)

[meson0731]

Homework Statement

summation 1 to infinity (2*4*6*8*...*(2n)) / (n!)

Homework Equations

The Attempt at a Solution

I'm a little confused on the "..." part... I used the ratio test and the numbers obviously cancel out, but I am not sure if the (2n+2)/(2n) part cancel out to just make (2n+2) on the top. This would make the limit equal to 2 and it would diverge. But I am not entirely sure if that's right...

 

[jgens]

It looks like you want to know if the following sum converges or diverges:
\sum_{n=1}^{\infty}\frac{2^n n!}{n!}=\sum_{n=1}^{\infty}2^n
That should be pretty easy.

 

[meson0731]

I don't see how you rewrote it like that. Would my method of using the ratio test work? And if so how exactly would the top partion cancel out.

 

[jgens]

I don't see how you rewrote it like that.

Try thinking about it. The ratio test works but is overkill for this problem.

Would my method of using the ratio test work? And if so how exactly would the top partion cancel out.

The ratio test works, but you messed up calculating it. (Hint: The ratio is 2).

 

[Robert1986]

Homework Statement

summation 1 to infinity (2*4*6*8*...*(2n)) / (n!)

Homework Equations

The Attempt at a Solution

I'm a little confused on the "..." part... I used the ratio test and the numbers obviously cancel out, but I am not sure if the (2n+2)/(2n) part cancel out to just make (2n+2) on the top. This would make the limit equal to 2 and it would diverge. But I am not entirely sure if that's right...

You said you were confused about the "..." part, so I just thought that I would mention that the n^{th} term in the sum is (2*4*\cdots*(2n))/n!.

Perhaps you knew this, but since you said you were confused by the "..." and jgens simplification, I thought you might be overlooking this.

 

[meson0731]

Try thinking about it. The ratio test works but is overkill for this problem.



The ratio test works, but you messed up calculating it. (Hint: The ratio is 2).

When i did the ratio test i did get 2 for the limit. Does this mean that I did it right?

 

[jgens]

When i did the ratio test i did get 2 for the limit. Does this mean that I did it right?

Depends. In the OP you wrote something about \lim_{n \to \infty}\frac{2n+2}{2n} = 2. In this case the limit is correct but the ratio is not. You should get \lim_{n \to \infty} 2 = 2.

 

[meson0731]

Depends. In the OP you wrote something about \lim_{n \to \infty}\frac{2n+2}{2n} = 2. In this case the limit is correct but the ratio is not. You should get \lim_{n \to \infty} 2 = 2.

I just realized that I wrote it wrong . I got lim as x > infinity of (2n+2)/(n+1) which is 2.

 

[jgens]

I just realized that I wrote it wrong . I got lim as x > infinity of (2n+2)/(n+1) which is 2.

Then you solved the problem correctly :)

 

[Robert1986]

Depends. In the OP you wrote something about \lim_{n \to \infty}\frac{2n+2}{2n} = 2. In this case the limit is correct but the ratio is not. You should get \lim_{n \to \infty} 2 = 2.

Wait, what? The limit doesn't look right to me...

\lim_{n \to \infty}\frac{2n+2}{2n} = \lim_{n \to \infty}\frac{2n}{2n} + \lim_{n \to \infty}\frac{2}{2n} = 1 + \lim_{n \to \infty}\frac{1}{n} = 1

Right?

 

[jgens]

Wait, what? The limit doesn't look right to me...

\lim_{n \to \infty}\frac{2n+2}{2n} = \lim_{n \to \infty}\frac{2n}{2n} + \lim_{n \to \infty}\frac{2}{2n} = 1 + \lim_{n \to \infty}\frac{1}{n} = 1

Right?

Yep. My bad.