tux
Decide convergence/divergence
from n=1 to infinity
(2^n+1)/(3^n1)



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anonymous
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converges for sure, and in fact i think the sum is not even 5

anonymous
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ignore the irrelavent 1 and +1 and use the root test, get it right away

tux
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It is hard to calculate using root test

anonymous
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not when the index is in the exponent, then it is easy

anonymous
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\[\sqrt[n]{2^n}=2\]!

tux
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What to do with +1 and 1?

tux
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Still have no idea how to do

Zarkon
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if you want...you can use the comparison test

tux
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Compare with (2/3)^n?

Zarkon
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sure...use the limit comparison test then.