(2n)/(n^n) does the infinte series converge?

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In summary, The conversation is about investigating the convergence of an infinite sum with a formula for its nth term given. The person has tried using ratio test and nth root test but was unsuccessful, and is now looking for other ideas. Another person suggests using Stirling's approximation. However, someone else points out that the limit using this method goes to infinity, indicating that the series diverges. The person then mentions that for the series to converge, the limit must be zero, but in this case, it is not.
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heshbon
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Homework Statement



Sorry for the missing summation signs but could anyone help me investigate the convergance of the following infinite sum with n'th term equal to : (2n)!/(n^n)


Homework Equations





The Attempt at a Solution



I have tried ratio test and n'th root test but failed.
Im not even sure if it passes the vanishing test

would appreciate any ideas. Thanks
 
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  • #2
Use Stirling's approximation.
 
  • #3
Well, using ratio test, i got that the limit goes to infinity, looks strange, but, i think that it diverges.
 
  • #4
using sterlings equationm plus nth root test i get a limit which tends to infinity, but i think you can only conclude something about convergance of the sseries if the limit is real...
 
  • #5
For the series to converge the limit must be zero, right? It's not.
 

1. What is the formula for calculating the convergence of (2n)/(n^n) infinite series?

The formula for calculating the convergence of an infinite series is lim n→∞ (Σ from k=1 to n of a_k). In the case of (2n)/(n^n), this would be lim n→∞ (2n)/(n^n).

2. How do you determine if the infinite series (2n)/(n^n) converges or diverges?

To determine the convergence or divergence of an infinite series, we need to use mathematical tests such as the ratio test, root test, or comparison test. These tests help us analyze the behavior of the terms in the series and determine if they approach a finite limit or not.

3. Can the value of (2n)/(n^n) infinite series be negative?

Yes, the value of an infinite series can be negative. The convergence or divergence of a series is determined by the behavior of its terms, not the overall value of the series.

4. Is there a specific value that (2n)/(n^n) infinite series converges to?

The value that an infinite series converges to is called the limit or sum of the series. For (2n)/(n^n), the value of the limit depends on the value of n. As n approaches infinity, the value of the limit will approach a certain value or go to infinity or negative infinity.

5. How does the convergence of (2n)/(n^n) infinite series relate to its rate of growth?

The convergence of an infinite series is related to its rate of growth. If the terms in the series approach a finite limit, then the series is said to converge, and its rate of growth would be considered slow. If the terms in the series do not approach a finite limit, then the series is said to diverge, and its rate of growth would be considered fast.

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