anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I will just continue from where you're stuck, assuming all you have done so far is right.
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty}{3^n \over 3^n-1}=1\] (since they both grow equally).
anonymous
  • anonymous
i do not understand how did u got one

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.