use the limit comparison test to determine whether sigma 2^n/3^n-1 converges or diverges. I solved the problem but i am stuck at this point lim n-->inf 3^n/3^n-1

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use the limit comparison test to determine whether sigma 2^n/3^n-1 converges or diverges. I solved the problem but i am stuck at this point lim n-->inf 3^n/3^n-1

Mathematics
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I will just continue from where you're stuck, assuming all you have done so far is right.
\[\lim_{n \rightarrow \infty}{3^n \over 3^n-1}=1\] (since they both grow equally).
i do not understand how did u got one

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Factor out 3^n from the numerator and denominator.
\[\frac{3^n}{3^n-1}=\frac{3^n}{3^n(1-\frac{1}{3^n})}=\frac{1}{1-\frac{1}{3^n}}\]
oh okay i see it
The limit as n approaches infinity is \[\frac{1}{1-0}=1\]
Hey Lucas :)
Hi Anwar :)
i have a question in a problem like this sigma ( 1/2n-1-1/2n) what do i compare it to
brb...one sec
k
Can you write that expression in the equation editor, because going off what you've written, your expression looks equal to -1.
?
\[\sum_{1}^{\infty} (1/ 2n-1-1/2n)\]
how do you write a fraction
There'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.
\[\sum_{1}^{\infty} \frac{1}{2n-1} -\frac{1}{2n} \]
Okay...now I can look at it :)
:)
i forgot we have to use the limit comparison test here
I think you'd be best to combine your fractions and use the limit comparison on the result. Your summand combines as\[\frac{1}{4n^2-2n}\]
This 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.
\[\frac{\frac{1}{4n^2-2n}}{\frac{1}{n^2}}=\frac{n^2}{4n^2-2n}=\frac{1}{4-2/n} \rightarrow \frac{1}{4}\]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.
hey i got the same things i did along with u thank u for ur help
you're welcome

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