# series convergent or divergent by ratio test

• Mar 5th 2009, 09:27 AM
oceanmd
series convergent or divergent by ratio test
Please determine whether the series is convergent or divergent by ratio test.
Series (2n-1)/2^(3n)
First I solved that lim a(n+1)/an =1, so the test fails, now I am rethinking it and if I do a more accurate division lim = 1/8, which makes the series convergent.

Thank you
• Mar 5th 2009, 09:37 AM
Krizalid
Quote:

Originally Posted by oceanmd

You can't, you just did it, there's no much here, you need to try other tests.

I'd just bound: $\displaystyle \frac{2n-1}{2^{3n}}<\frac{2^{n}}{2^{3n}}=\left( \frac{1}{4} \right)^{n},$ whereat your series converges by direct comparison test with a geometric one.
• Mar 5th 2009, 09:52 AM
oceanmd
where is the mistake in the calculation below
Then, please explain where is the mistake in the calculation below
Lim [2(n+1)-1/2^(3(n+1))] / [(2n-1)/2^(3n)] = lim (2n+1) / 2^(3n) 2^3) times 2^(3n)/92n-1) = lim (2n+1)/(16n-8)= 1/8

I substituted 2^(3n+3) by 2^(3n) times 2^3
• Mar 5th 2009, 09:57 AM
Krizalid
Very hard to read, but let's compute $\displaystyle \frac{a_{n+1}}{a_n}:$

$\displaystyle \frac{2(n+1)-1}{2^{3(n+1)}}\cdot \frac{2^{3n}}{2n-1}=\frac{2n+1}{2n-1}\cdot \frac{2^{3n}}{2^{3n}\cdot 8}\to \frac{1}{8},$ as $\displaystyle n\to\infty,$ and yes, you're correct, it converges by applying the ratio test. (On my previous post you said the limit equals $\displaystyle 1,$ thus I assumed you were right.)
• Mar 5th 2009, 11:00 AM
oceanmd
Thank you a lot
Thank you very much.

When I said lim = 1, this is what I came up with, but then I thought, wait a minute, this is not right.

Thank you again