# Uniform convergens and continuity on R

1. May 10, 2006

### Hummingbird25

Hello people,

I'm tasked with showing the following:

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

(1) show that it converges Uniformly $$f_n(x) :\mathbb{R} \rightarrow \mathbb{R}$$.

(2) Next show the function

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

is continious on $$\mathbb{R}$$

(1) Suppose $$f_n = \frac{1}{x^2 + n^2}$$,

Then $$f_n$$ is uniformly convergens if

$$sup _{x \in \mathbb{R}} |f_n(x) - x|$$. Now

$$_{sup} _{x \in \mathbb{R}} |f_n(x) - x| = |\frac{1}{x^2 - n^2} - x| = _{sup} _{x \in \mathbb{R}} \frac{1}{x^2 - n^2}$$

The deriative of f_n(x) is non-negative on $$\mathbb{R}$$, so its increasing and is hence maximumized at $$x = \mathbb{R}$$. So the supremum is $$1/n^2$$. This does tend to zero as $$n \rightarrow \infty$$. So therefore it converge Uniformly.

Am I on the right track here? If yes any hints on how to prove the continuety ?

I know that its something to do with:

\integral_{1} ^{\infty} 1/x^2 + n^2 dx = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}

Sincerely Hummingbird25

Last edited: May 10, 2006
2. May 10, 2006

### matt grime

you've not defined f_n

You start an 'if' statement and then do not complete it with a condition about anything.

3. May 10, 2006

### Hummingbird25

Hi :)

f_n: \mathbb{R} \rightarrow \mathb{R}

f_n(x) should have said f_n

There is one sup to much sorry.

Sincerely

Hummingbird25

Last edited: May 10, 2006