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anonymous
 5 years ago
use the limit comparison test to determine whether sigma 2^n/3^n1 converges or diverges. I solved the problem but i am stuck at this point lim n>inf 3^n/3^n1
anonymous
 5 years ago
use the limit comparison test to determine whether sigma 2^n/3^n1 converges or diverges. I solved the problem but i am stuck at this point lim n>inf 3^n/3^n1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will just continue from where you're stuck, assuming all you have done so far is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty}{3^n \over 3^n1}=1\] (since they both grow equally).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i do not understand how did u got one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Factor out 3^n from the numerator and denominator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{3^n}{3^n1}=\frac{3^n}{3^n(1\frac{1}{3^n})}=\frac{1}{1\frac{1}{3^n}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limit as n approaches infinity is \[\frac{1}{10}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a question in a problem like this sigma ( 1/2n11/2n) what do i compare it to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you write that expression in the equation editor, because going off what you've written, your expression looks equal to 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} (1/ 2n11/2n)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you write a fraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a box in the equation editor, or you can type, frac{}{} and put your numerator stuff in the first box, denominator stuff in the second.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \frac{1}{2n1} \frac{1}{2n} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...now I can look at it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i forgot we have to use the limit comparison test here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you'd be best to combine your fractions and use the limit comparison on the result. Your summand combines as\[\frac{1}{4n^22n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is asymptotically equivalent to 1/(4n^2), which means you can eyeball the fact you could compare it to 1/n^2 for limit comparison.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\frac{1}{4n^22n}}{\frac{1}{n^2}}=\frac{n^2}{4n^22n}=\frac{1}{42/n} \rightarrow \frac{1}{4}\]as n goes to infinity. The limit is positive and finite, so by the limit comparison test, since Sum(1/n^2) converges, your series does too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey i got the same things i did along with u thank u for ur help
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