Ratio test for math convergence

[DotKite]

Homework Statement

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

Homework Equations

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.

 

[Dick]

Homework Statement

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

Homework Equations

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.

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.

 

[DotKite]

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.

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 into 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.

 

[haruspex]

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

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.

 

[haruspex]

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

 

[Quinzio]

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

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

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

Then I plug it back into the original function,

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

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
$\frac{x^2}{(1+x^2)^n}<\frac{1+x^2}{(1+x^2)^n}=\frac{1}{(1+x^2)^{n-1}}$

 

[haruspex]

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.

Nicer, but not convergent.

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

That won't give uniform convergence.

 

[Quinzio]

Hi haruspex,

It will converge in all reals but zero.

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

 

[haruspex]

It will converge in all reals but zero.

The expression I passed that comment on did not involve x, so I don't understand your answer.

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

Just because an attempt to prove uniform convergence fails, you cannot conclude that it is not uniformly convergent. (It is.)

 

[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.

 

[haruspex]

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.

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.

 

[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.

 

[Dick]

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.

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.

 

[DotKite]

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.

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.