anonymous
  • anonymous
Power Series: sigma n=1 n-> infinity ((n(n+1)^n)/4^n) so far ive gotten IxI < (1/4) how do i find the interval of convergence and radius of convergence?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
you dont have x in your summation, how can you get |x| < 1/4?
anonymous
  • anonymous
woops sigma n=1 n-> infinity ((n(x+1)^n)/4^n) IxI < (1/4)
anonymous
  • anonymous
hello again btw :]

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anonymous
  • anonymous
great :)
anonymous
  • anonymous
by using the root test, you will get the limit (n^(1/n))[(x+1)/4]
anonymous
  • anonymous
since lim (n^(1/n)) = 1, for the series to converge, you get |x+1| < 4
anonymous
  • anonymous
oh okay hmm okay now how would I find the intervals of convergence?
anonymous
  • anonymous
the radius of convergence is 4, and it is around -1. So the interval we are interested is (-5,3)
anonymous
  • anonymous
okay so we plug in what makes it zero which in this case is -1, and how did you get the interval?
anonymous
  • anonymous
by solving |x+1| < 4. -4 < x+1 < 4 -5 < x < 3
anonymous
  • anonymous
Ohhh! xD thanks
anonymous
  • anonymous
So check the convergence of the series at the end points
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
how would I know to apply the root test? ratio?

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