## anonymous 5 years ago 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

1. anonymous

I will just continue from where you're stuck, assuming all you have done so far is right.

2. anonymous

$\lim_{n \rightarrow \infty}{3^n \over 3^n-1}=1$ (since they both grow equally).

3. anonymous

i do not understand how did u got one

4. anonymous

Factor out 3^n from the numerator and denominator.

5. anonymous

$\frac{3^n}{3^n-1}=\frac{3^n}{3^n(1-\frac{1}{3^n})}=\frac{1}{1-\frac{1}{3^n}}$

6. anonymous

oh okay i see it

7. anonymous

The limit as n approaches infinity is $\frac{1}{1-0}=1$

8. anonymous

Hey Lucas :)

9. anonymous

Hi Anwar :)

10. anonymous

i have a question in a problem like this sigma ( 1/2n-1-1/2n) what do i compare it to

11. anonymous

brb...one sec

12. anonymous

k

13. anonymous

Can you write that expression in the equation editor, because going off what you've written, your expression looks equal to -1.

14. anonymous

?

15. anonymous

$\sum_{1}^{\infty} (1/ 2n-1-1/2n)$

16. anonymous

how do you write a fraction

17. 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.

18. anonymous

$\sum_{1}^{\infty} \frac{1}{2n-1} -\frac{1}{2n}$

19. anonymous

Okay...now I can look at it :)

20. anonymous

:)

21. anonymous

i forgot we have to use the limit comparison test here

22. anonymous

I think you'd be best to combine your fractions and use the limit comparison on the result. Your summand combines as$\frac{1}{4n^2-2n}$

23. 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.

24. 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.

25. anonymous

hey i got the same things i did along with u thank u for ur help

26. anonymous

you're welcome