# Show convergence of this series

• dipole
In summary: So, I see you are using the ratio test. That is the first thing I would try.However, since you are dealing with a log, the integral test should workIn summary, the person is seeking help in showing that the infinite series given by Sigma (1-n*ln((2n+1)/(2n-1))) converges. They have tried using the ratio test and L'Hopital's rule, but it resulted in an indeterminate answer. Another person suggests trying the integral test by splitting the log into the difference of two logs and using integration by parts. The conversation then leads to discussing the use of small x values and upper and lower bounds to determine the convergence of the series. The expert summarizer concludes
dipole

## Homework Statement

All I want to show is that the following infinite series converges,

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

## Homework Equations

Various series tests...

## 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?

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.

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...

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.

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?

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?

dipole said:

## Homework Statement

All I want to show is that the following infinite series converges,

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

## Homework Equations

Various series tests...

## 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?

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.

dipole said:
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?

it is rendering now.

## What is a series?

A series is a sum of an infinite sequence of numbers.

## What does it mean for a series to converge?

A series converges if the sum of its terms approaches a finite number as the number of terms increases.

## What is the difference between a convergent and a divergent series?

A convergent series has a finite sum, while a divergent series has an infinite or undefined sum.

## What are some common methods for showing convergence of a series?

Some common methods include the comparison test, the ratio test, and the integral test.

## Why is it important to determine the convergence of a series?

Determining the convergence of a series allows us to determine whether the sum of the terms will approach a finite value, which can have practical applications in various fields of science and math.

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