Power Series- radius of convergence

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SUMMARY

The radius of convergence for the power series \(\sum^{\infty}_{n=1}\frac{n!x^n}{n^n}\) is determined using the Ratio Test. The limit evaluated is \(\left(\frac{n}{n+1}\right)^n\), which approaches \(e^{-1}\) as \(n\) approaches infinity. Consequently, the radius of convergence is \(e\), as the series converges when \(|x| < e\). The initial confusion regarding the application of L'Hôpital's rule was clarified, emphasizing that it was unnecessary in this context.

PREREQUISITES
  • Understanding of power series and convergence
  • Familiarity with the Ratio Test for series convergence
  • Basic knowledge of limits and logarithmic functions
  • Experience with factorial notation and its implications in series
NEXT STEPS
  • Study the application of the Ratio Test in various power series
  • Learn about the properties of factorial growth in series
  • Explore the concept of convergence radii in more complex series
  • Investigate the use of L'Hôpital's rule in limit evaluations
USEFUL FOR

Mathematics students, educators, and anyone studying series convergence in calculus or advanced mathematics courses.

Roni1985
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Homework Statement


determine the radius of convergence of the given power series

[tex]\sum[/tex][tex]^{inf}_{n=1}[/tex][tex]\frac{n!x^n}{n^n}[/tex]

Homework Equations





The Attempt at a Solution


I did the ratio test
then I had to take the 'ln'
but, my answer is this
|e|<1 for the series to converge.
It never happens but according to the answers the radius is 'e'.
 
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oh I think I got it, I didn't have to lhopital lnx in the middle ...
:\
 
Applying the Ratio Test, we have the sequence

[tex]\frac{(n+1)n^n}{(n+1)^{(n+1)}} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n[/tex]

Taking the limit as n goes to infinity... this might look like a familiar limit. Then recall that the Radius of convergence is the reciprocal of this limit.
 

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