Thread: series convergent or divergent by ratio test

1. 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

2. 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.

3. 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

4. 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.)

5. 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