anonymous
  • anonymous
how do you know (-1)^n(1+4^n/1+3^n) is a geometric series?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[(-1)^n\frac{1+4^n}{1+3^n}\] ? is that right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
idk how to prove that its a geometric series

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anonymous
  • anonymous
I am not sure if it is even a geometric series what do you think @zarkon
anonymous
  • anonymous
well they used a geometric series to compare it with, the answer diverges
anonymous
  • anonymous
oh, that's different; it is called comparison test; this does not have to be geometric series for that
anonymous
  • anonymous
let me know if you are still here
Zarkon
  • Zarkon
you can also use the divergence test. The limit of the sequence is not zero...therefore the series diverges
Zarkon
  • Zarkon
and imranmeah91is correct....it is not geometric.
anonymous
  • anonymous
|dw:1332465387707:dw| this is what they had used on a previous problem, so i figured id try the same method, but i dont knw how they used a geometric series to compare it with to get the answer, that is what that puzzles me, any ideas?
Zarkon
  • Zarkon
\[\frac{1+4^n}{1+3^n}>\frac{4^n}{1+3^n}>\frac{4^n}{3^n+3^n}=\frac{1}{2}\frac{4^n}{3^n}=\frac{1}{2}\left(\frac{4}{3}\right)^{n}\to \infty \text{ as }n\to\infty\] thus by the divergence test the series does not converge.
anonymous
  • anonymous
where did u get the divergence test from i couldnt find it in my calc book
Zarkon
  • Zarkon
http://mathworld.wolfram.com/DivergenceTests.html
anonymous
  • anonymous
yeah, but... how did u knw to use a geometric series to compare it with, was it cuz u recognized the pattern or something
anonymous
  • anonymous
wait nvm im stupid i see it now
Zarkon
  • Zarkon
I just transformed it into something I could easly take the limit of.
anonymous
  • anonymous
how r u able to compare it with 4^n/3^n+3^n ? are u allowed to do that?

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