tux
  • tux
Decide convergence/divergence from n=1 to infinity (2^n+1)/(3^n-1)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
converges for sure, and in fact i think the sum is not even 5
anonymous
  • anonymous
ignore the irrelavent -1 and +1 and use the root test, get it right away
tux
  • tux
It is hard to calculate using root test

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anonymous
  • anonymous
not when the index is in the exponent, then it is easy
anonymous
  • anonymous
\[\sqrt[n]{2^n}=2\]!
tux
  • tux
What to do with +1 and -1?
tux
  • tux
Still have no idea how to do
Zarkon
  • Zarkon
if you want...you can use the comparison test
tux
  • tux
Compare with (2/3)^n?
Zarkon
  • Zarkon
sure...use the limit comparison test then.

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