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

  • Thread starter meson0731
  • Start date
For some reason I thought it was \lim_{n \to \infty}\frac{2n+2}{2n} = 2 + \lim_{n \to \infty}\frac{2}{2n} = 2 and was really confused. I should know better than to do math late at night lol. Thanks for catching that!In summary, the sum \sum_{n=1}^{\infty}\frac{2^n n!}{n!} converges to 2, which was found using the ratio test. However, the limit was incorrectly calculated in the original attempt, but was later corrected to 1 by another user.
  • #1
meson0731
14
0

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...
 
Physics news on Phys.org
  • #2
It looks like you want to know if the following sum converges or diverges:
[tex]\sum_{n=1}^{\infty}\frac{2^n n!}{n!}=\sum_{n=1}^{\infty}2^n[/tex]
That should be pretty easy.
 
  • #3
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.
 
  • #4
meson0731 said:
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).
 
  • #5
meson0731 said:

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 [itex]n^{th}[/itex] term in the sum is [itex](2*4*\cdots*(2n))/n![/itex].

Perhaps you knew this, but since you said you were confused by the "..." and jgens simplification, I thought you might be overlooking this.
 
  • #6
jgens said:
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?
 
  • #7
meson0731 said:
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 [itex]\lim_{n \to \infty}\frac{2n+2}{2n} = 2[/itex]. In this case the limit is correct but the ratio is not. You should get [itex]\lim_{n \to \infty} 2 = 2[/itex].
 
  • #8
jgens said:
Depends. In the OP you wrote something about [itex]\lim_{n \to \infty}\frac{2n+2}{2n} = 2[/itex]. In this case the limit is correct but the ratio is not. You should get [itex]\lim_{n \to \infty} 2 = 2[/itex].

I just realized that I wrote it wrong :rolleyes: . I got lim as x > infinity of (2n+2)/(n+1) which is 2.
 
  • #9
meson0731 said:
I just realized that I wrote it wrong :rolleyes: . I got lim as x > infinity of (2n+2)/(n+1) which is 2.

Then you solved the problem correctly :)
 
  • #10
jgens said:
Depends. In the OP you wrote something about [itex]\lim_{n \to \infty}\frac{2n+2}{2n} = 2[/itex]. In this case the limit is correct but the ratio is not. You should get [itex]\lim_{n \to \infty} 2 = 2[/itex].

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

[tex]\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[/tex]

Right?
 
  • #11
Robert1986 said:
Wait, what? The limit doesn't look right to me...

[tex]\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[/tex]

Right?

Yep. My bad.
 

1. What is the formula for the sum of the series 2*4*6*8*...*(2n) / (n!)?

The formula for the sum of this series is ∑ (2n) / (n!), where n starts at 1 and goes to infinity.

2. What is the convergence or divergence of this series?

This series is a divergent series, meaning that the sum of the series does not have a finite value.

3. How do you calculate the partial sum of this series?

The partial sum can be calculated by plugging in a specific value for n in the formula ∑ (2n) / (n!). For example, the partial sum for n = 5 would be 2*4*6*8*10 / 5! = 3.2.

4. What is the significance of the factorial in the denominator of the formula?

The factorial in the denominator is used to account for the increasing denominator in each term of the series. This ensures that the series does not approach infinity as n increases.

5. Can this series be used in real-life applications?

Yes, this series can be used in various mathematical and scientific calculations, such as probability and statistics, to determine the likelihood of certain events occurring.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
937
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
7K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
966
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Back
Top