# Show convergence of this series

1. Oct 3, 2013

### dipole

1. The problem statement, all variables and given/known data
All I want to show is that the following infinite series converges,

$\Sigma_{n=1}^{\inf} = \bigg(1 - n\ln\big(\frac{2n+1}{2n-1}\big)\bigg)$

2. Relevant equations

Various series tests...

3. The attempt at a solution

I tried doing a ratio test, after applying L'Hopital's rule, but I simply get 1, which is indeterminate... can anyone help me get this thing solved? Maybe suggest the proper test to use?

2. Oct 3, 2013

### brmath

Try the integral test. If you split your log into the difference of two logs, you can use integration by parts . Then put all 3 pieces over the lowest common denominator, and see what you have. Don't forget the constant term that pops out of the integration by parts.

The series converges if and only if the integral converges.

3. Oct 3, 2013

### dipole

I don't think that helps me, because integrating I get,

$I = \frac{1}{8}\bigg(4x+(4x^2-1)\ln\big(\frac{2x-1}{2x+1}\big)\bigg) |_1^{\inf}$

which seems like a very difficult limit to evaluate...

4. Oct 3, 2013

### brmath

Gee, that's not what I got. How did you do the integral?

BTW, are you able to write your answers in Latex? The Physics Forum site has a very clear discussion about how to do so. I keep it up in one tab all the time.

5. Oct 4, 2013

### dipole

I actually used Mathematica, because I didn't feel like wasting more time if it didn't work, so the result should be correct.

Also, I don't understand what you mean, it is typed up in LaTex, are you not seeing the rendered equations?

6. Oct 4, 2013

### haruspex

I can see how to reduce it to showing that $\Sigma (1-n \ln(1+1/n))$ and $\Sigma (1+n \ln(1-1/n))$ converge. Does that help?

7. Oct 4, 2013

### Ray Vickson

You can set 2n = 1/x in
$$t_n = 1 - n \ln \left(\frac{2n+1}{2n-1} \right)$$ and expand in a series for small x, to get nice upper and lower bounds on $t_n$ for large n.

8. Oct 4, 2013

### brmath

it is rendering now.