# Ratio test for math convergence

1. Jun 8, 2013

### DotKite

1. The problem statement, all variables and given/known data

show $\sum \frac{x^{2}}{(1+x^{2})^{n}}$ converges uniformly on R

2. Relevant equations

3. The attempt at a solution

I know by ratio test it is absolutely convergent for all x in R.

I am guessing you use m-test. However I do not really understand how m-test works. It is the end of the quarter and my professor pretty much trail blazed through this last chapter in order to finish on time.

2. Jun 8, 2013

### Dick

The first step is to find the $M_n$ in the m-test. Find the maximal value of each function in your sum. Use calculus.

3. Jun 8, 2013

### DotKite

Ok, so I let $f_n$(x) = $\frac{x^{2}}{(1+x^{2})^{n}}$

and took the derivative,

$f_n$'(x) = $\frac{2x}{(x^{2}+1)^{n+1}} * [1 - (n-1)x^{2}]$

when i set it equal to zero and solve I get x = ± $\frac{1}{(n-1)^{\frac{1}{2}}}$

Then I plug it back in to the original function,

$f_n$ $\left ( \frac{1}{(n-1)^{\frac{1}{2}}} \right )$ = $\frac{1}{(n-1)(1+\frac{1}{n})^{n}}$

before I try to figure out if $M_n$ = $\frac{1}{(n-1)(1+\frac{1}{n})^{n}}$ is a convergent series please tell me if I am completely wrong already.

4. Jun 9, 2013

### haruspex

That's not what I get, but the difference shouldn't matter much for the convergence or otherwise. It's roughly 1/(n e) either way, which clearly doesn't converge, so you'll need a more sensitive test.

5. Jun 9, 2013

### haruspex

The series is just a GP. If the limit is S and the sum of the first r terms is Sr, consider the error term, S-Sr.

6. Jun 9, 2013

### Quinzio

First of all the thesis that $f_n(x)$ in uniformly convergen on $\mathbb{R}$ is not to be taken for sure.

Those are not the only values by which $f'(x)=0$

Something went wrong during this step. You should get:
$\frac{\frac{1}{1-n}}{\left(1+\frac{1}{1-n}\right)^n}$
and with some easy manipulation you get something nice.

another nicer way is to see that
$\frac{x^2}{(1+x^2)^n}<\frac{1+x^2}{(1+x^2)^n}=\frac{1}{(1+x^2)^{n-1}}$

7. Jun 9, 2013

### haruspex

Nicer, but not convergent.
That won't give uniform convergence.

8. Jun 9, 2013

### Quinzio

Hi haruspex,

It will converge in all reals but zero.

That's why I'm saying the thesis is not true.

Last edited: Jun 9, 2013
9. Jun 9, 2013

### haruspex

The expression I passed that comment on did not involve x, so I don't understand your answer.
Just because an attempt to prove uniform convergence fails, you cannot conclude that it is not uniformly convergent. (It is.)

10. Jun 9, 2013

### Quinzio

OK ok.
I only wanted to give an easy way for uniform convergence for $x \in [-\infty,0),(0,\infty]$.
Then one should treat the zero as a special case.

11. Jun 9, 2013

### haruspex

You're not going to show uniform convergence by that method even on $(0,\infty]$. You would be able to show it for $(\delta,\infty]$ for any δ>0, but that's not the same.

12. Jun 11, 2013

### DotKite

decided to go about a different way. I wrote out the nth partial sum

$s_{1}=\frac{x^{2}}{1+x^{2}}$

$s_{2}=\frac{x^{2}}{\left ( 1+x^{2} \right )^{2}}$
.
.
.

$s_{n} = \frac{x^{2}}{1+x^{2}} + \frac{x^{2}}{\left ( 1+x^{2} \right )^{2}} + ... + \frac{x^{2}}{\left ( 1+x^{2} \right )^{n}}$ $= x^{2}\left ( \frac{1}{1+x^{2}} +...+\frac{1}{\left ( 1+x^{2} \right )^{n}} \right )$

notice

$\left ( \frac{1}{1+x^{2}} +...+\frac{1}{\left ( 1+x^{2} \right )^{n}} \right )$

is a geometric series

thus $s_{n} = \left (x^{2} + 1 \right ) \left [ 1 - \left ( \frac{1}{1+x^{2}} \right )^{n} \right ]$

my question: is that uniformly convergent? I am kind of confused on how to show a sequence is uniformly convergent. I know if I am able to show that then it means the series is uniformly convergent.

Last edited: Jun 11, 2013
13. Jun 11, 2013

### Dick

Now that's a good idea! But I don't think you've got the algebra quite right. Try it again. Consider that the behavior at x=0 might be quite different than elsewhere and you might find it's pretty easy to show that it CAN'T be uniformly convergent.

14. Jun 11, 2013

### DotKite

maybe it's just late, and my brain is fried. I now end up with

$\frac{1-\left ( \frac{1}{1 + x^{2}} \right )^{n}}{1 + x^{2}}$

doesn't seem right.

Last edited: Jun 11, 2013