Convergence and Continuity of a Series of Functions

[letmeknow]

Homework Statement

Show that the series,

$$\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}$$

defines a continuous function f on the domain of convergence. What is this domain? In addition, write a series representation of ( f ' ) and determine the domain of convergence of this series to ( f ' ).

Homework Equations

The Attempt at a Solution

I need abit of help with this problem. If somebody could point me in the direction I would be very happy.

It looks to me that the series in question might be smaller than the series 1/ (k^2) and therefore converges on a domain of all real numbers.

I had a question about the wording, "the series defines a continuous function f on the domain of convergence". Does this mean that I am looking for the function that this series uniformly or pointwise converges to?

 

[Office_Shredder]

I had a question about the wording, "the series defines a continuous function f on the domain of convergence". Does this mean that I am looking for the function that this series uniformly or pointwise converges to?

In a manner of speaking, but you can consider it more directly. If, given a value of x, the sum converges, then we can define f(x) = the value of the summation for x, and this is the function they're referring to.

 

[letmeknow]

I have that for,

$$x=0 , \sum_{n=1}^{\infty}\frac{1}{n^2}$$

$$x=1 , \sum_{n=1}^{\infty}\frac{1}{1+n^2}$$

$$x=2 , \sum_{n=1}^{\infty}\frac{1}{4+n^2}$$

I know that these converge to 0. So no matter what are x values, this series converges. I would say that the domain of convergence is all of R. Am I right with that?

So did the series define the function f(x)=0. I am having trouble understanding what this means.

 

[letmeknow]

Ok, so this series converges to some function f and I must find what the interval of x is in which this happens.

Second, if we take the derivative of f, say ( f ' ), the same series also converges to ( f ' ) but on a different (or same) interval ?

 

[lanedance]

why not have a look at the term by term differentiation of f? i think it asks for this is the question

 

[HallsofIvy]

I have that for,

$$x=0 , \sum_{n=1}^{\infty}\frac{1}{n^2}$$

$$x=1 , \sum_{n=1}^{\infty}\frac{1}{1+n^2}$$

$$x=2 , \sum_{n=1}^{\infty}\frac{1}{4+n^2}$$

I know that these converge to 0. So no matter what are x values, this series converges. I would say that the domain of convergence is all of R. Am I right with that?

So did the series define the function f(x)=0. I am having trouble understanding what this means.

Surelly you know that showing that a function converges for 3 values of x does NOT show that it converges for ALL x. Also, it is clear that those DO NOT "convert to 0". The very first one starts 1+ 1/4+ 1/9 and has only positive terms.

 

[letmeknow]

why not have a look at the term by term differentiation of f? i think it asks for this is the question

I have a theorem that says that the radius of convergence of a power series is the same as the radius of convergence of the term by term differentiation of the power series.

I didn't know if this was a power series or not. I cannot see how to separate the a_n and the x^n if it is, and if it isn't does the result apply also to series like the one I have brought up?

Surelly you know that showing that a function converges for 3 values of x does NOT show that it converges for ALL x. Also, it is clear that those DO NOT "convert to 0". The very first one starts 1+ 1/4+ 1/9 and has only positive terms.

I am confused about this though. I though that I could show that the summation for 1/k^2 converges by the ratio test by proving that for n large enough the terms of the ratio are all less than 1. Then for any other x in R, x^2 is positive and would be adding more to the denominator, thus making it even smaller and thus would converge also.

What am I confusing here? I understand that we are adding positive terms now though, and the series all converge to something that is not necessarily 0.

 

[letmeknow]

I understand the problem better now. I don't think I need to find the equation of the function that it converges to. (It would look like a bell curve though, I'm guessing)

I used the ratio test (with the x's in there) and determined that the series converges for all x in R. (R = infinity). I then did a term by term differentiation and performed the ratio test on that to get the same radius of convergence (all x in R).

I've learned a couple things here about the function. Is there anything else I am missing? (Assuming that doing the 2 ratio tests were valid?)

 

[lanedance]

not sure what your lecturer wants... but you have shown both f & f' converge on a radius of covergence, and determined this radius

Reading the question I think you might still need to show continuity on the domain of convergence... maybe a quick epsilon delta?

 

[letmeknow]

not sure what your lecturer wants... but you have shown both f & f' converge on a radius of covergence, and determined this radius

Reading the question I think you might still need to show continuity on the domain of convergence... maybe a quick epsilon delta?

I was wondering if I could do that. Do you mean something like this?

$$| \sum_{n=1}^{\infty}\frac{1}{x^2+n^2}- \sum_{n=1}^{\infty}\frac{1}{a^2+n^2}|$$

I tried this to no avail after I combined? the series into one. I found one |x-a| in there but it seemed like nothing else.