Interval of convergence problem

In summary, the original series is the sum from n=1 to infinity of (n!*x^n)/(n^n) where x is a positive real number and Stirling's approximation is used to get e's out of it. The interval of convergence is (-e < x < e). But now I need to test the endpoints, which means I need to find if the following two series converge: sum from n=1 to infinity of (n!*e^n)/(n^n) and sum from n=1 to infinity of (n!*(-e)^n)/(n^n). I started with the first one, because the second is just an alternating version of it (right
  • #1
sinas
15
0
The original series is the sum from n=1 to infinity of (n!*x^n)/(n^n)

I used a ratio test to find that the interval of convergence is -e < x < e

But now I need to test the endpoints, which means I need to find if the following two series converge:

sum from n=1 to infinity of (n!*e^n)/(n^n)
sum from n=1 to infinity of (n!*(-e)^n)/(n^n)

I started with the first one, because the second is just an alternating version of it (right?) so if I proved absolute convergence I wouldn't have to do the second one as well. Here is where I am having problems though, I can't figure out a test that would prove divergence or convergence for the first one. I tried a ratio test, but I get an answer of 1 (inconclusive), which makes sense since I used a ratio to find the interval of convergence. Can someone point me in the right direction?
 
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  • #2
Did you try applying Stirling's approximation to this one too?
 
  • #3
Don't know what that is but I'll look it up in a sec...
 
  • #4
Ahh that works out nicely...

n!=~n^n*e^-n*(2*n*pi)^(1/2)

which leaves me with...

sum from n=1 to infinity of (2*n*pi)^(1/2)
and
sum from n=1 to infinity of (-1)^n*(2*n*pi)^(1/2)

and in both cases the "infiniteth" term doesn't go to zero, so both diverge... is my reasoning sound?
 
  • #5
Is there a way to show that the error in stirling's approximation goes to zero as n goes to infinity?
 
  • #6
I had assumed Stirling's approximation when applying the ratio test -- that's usually the only way I get e's out of it.

There are indeed more detailed forms of Stirling's approximation, but now that you know it diverges, maybe there's an easier way to show the terms don't converge to zero?
 
  • #7
Oh, BTW, the error doesn't go to zero, but the relative error does. All you really care about, though, is getting a lower bound, so it doesn't matter if there's any sort of convergence, just that you have a useful inequality.

Mathworld or Wikipedia probably have it.
 
  • #8
kk thanks :smile:
 

What is the "Interval of Convergence" problem?

The "Interval of Convergence" problem refers to finding the range of input values for a given function where the power series representation of the function will converge to the actual value of the function. This is a common problem in calculus and real analysis.

What is the significance of the "Interval of Convergence"?

Knowing the interval of convergence allows us to determine the accuracy of the power series approximation of a function. It also helps us determine if a given function can be represented as a power series at all.

How is the "Interval of Convergence" calculated?

The "Interval of Convergence" is typically found by using the ratio or root test to determine the values of x for which the series converges. This is done by considering the limit of the ratio or root of successive terms in the series and using the resulting value to find the interval of convergence.

What happens if the "Interval of Convergence" is infinite?

If the "Interval of Convergence" is infinite, it means that the power series representation of the function will converge for all values of x. This is also known as a series with a "radius of convergence" of infinity.

What are some common mistakes made when finding the "Interval of Convergence"?

One common mistake is using the ratio or root test incorrectly. Another mistake is not checking the endpoints of the interval to see if the series converges or not. It is also important to remember that the "Interval of Convergence" may not be a continuous interval, and may have gaps or discontinuities.

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