anonymous
  • anonymous
Find the radius/interval of convergence for (series posted below) . . .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{n=1}^{\infty} n^nx^n\] I attempted the root test, but that told me the series was divergent . . . unless I am applying the root test incorrectly??
amistre64
  • amistre64
is that n^n * x^n ?
anonymous
  • anonymous
Yes. I'm sorry :P

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amistre64
  • amistre64
it this tiny little laptop screen, hard to make stuff out :)
amistre64
  • amistre64
if i were to take a stab at this: \[\frac{n^nx^n}{(n-1)^{n-1}x^{n-1}}\] \[\frac{n^nx}{(n-1)^{n-1}}\] \[|x|\lim_{n\to inf}\frac{n^n}{(n-1)^{n-1}}\] \[|x|\lim_{n\to inf}\frac{n^n}{(n-1)^{n}(n-1)^{-1}}\] \[|x|\lim_{n\to inf}\frac{n^n(n-1)}{(n-1)^{n}}\] \[|x|\lim_{n\to inf}\frac{n^{n+1}-n^n}{(n-1)^{n}}\] yeah, i got no idea where this is going yet
amistre64
  • amistre64
i think the top goes to zero as the bottom goes to inf
Zarkon
  • Zarkon
the series only converges when x=0...I'll let you think about that :)
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=limit+%28n%5En%29%2F%28n-1%29%5E%28n-1%29+n+to+inf yeah, for the moment all i get is divergent
anonymous
  • anonymous
Oh, okay, so the ratio test worked . . . and if it's 0/infinity, then . . . that's 0? So the radius/interval would both be 0?
amistre64
  • amistre64
the wolf says it diverges
anonymous
  • anonymous
Okay. :< Thanks to both of y'all :)
Zarkon
  • Zarkon
it diverges everywhere except at x=0

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