Finding Radius of Convergence: Calculating (1/n!)x^(n!)

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SUMMARY

The radius of convergence for the series (1/n!)x^(n!) is determined to be 1, with the interval of convergence being [-1, 1]. The analysis confirms that the series converges for |x| < 1 using the comparison test, while it diverges for |x| > 1. At the endpoints x = 1 and x = -1, the series also converges. The conclusion is that the series converges for |x| ≤ 1.

PREREQUISITES
  • Understanding of series convergence tests, specifically the comparison test.
  • Familiarity with factorial notation and its implications in series.
  • Knowledge of absolute values in the context of convergence.
  • Basic concepts of interval notation and radius of convergence.
NEXT STEPS
  • Study the Ratio Test for series convergence and its applicability.
  • Explore the concept of power series and their convergence properties.
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garyljc
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Hey ,
I was wondering if anyone could help me out with this question regarding calculating the radius of convergence of the infinity series of (1/n!)x^(n!)

This is my work

First we consider when abs(x) < 1
then we have 0 <= abs(x^n!) <= abs(x^n)
so we know that the series converges whenever abs(x) < 1 , so by the comparison test, we conclude that the series converges for abs(x) < 1

Now consider abs(x) > 1
we have 0<= abs (x) <= abs(x^n!)
so the series by comparison test diverges for abs(x) > 1

After that , I consider what happen at x=1 and x=-1 , it turned out that they converges as well .
So my radius of convergence is [-1,1] and the series converges for abs(x) <= 1

is this approach correct ? because i thought that for question like this , ratio test is not a good approach.
 
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I don't see anything wrong, except that your radius of convergence, R, is 1. The interval of convergence is [-1, 1], and the radius of this interval is 1.
 
Alright thanks a lot =D .
 

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