Is the Sum-Function of a Convergent Series Continuous on $\mathbb{R}$?

[Math_Frank]

Hello

I have this question here which has puzzled me.

Given a series

$$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}$$

Show that the series converge for every $$y \in \mathbb{R}$$

By the test of comparison

$$\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}$$

Since its know that

$$\sum \limit_{n=0} ^{\infty} \frac{1}{n^2}$$ converge, then the series $$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}$$ converge for every $$y \in \mathbb{R}$$

Second show that the series converge Uniformt on $$\mathbb{R}$$

Again since

$$\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}$$

and since $$\sum \limit_{n=0} ^{\infty} \frac{1}{n^2}$$ converge.

Then by Weinstrass M-Test, then series Converge Uniformt on $$\mathrm{R}$$

Third show that the sum-function

$$f: \mathbb{R} \rightarrow \mathbb{R}$$

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}$$

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

Can I conclude here that since the series converge Uniformly on R, then its sum-functions is continuous on $$\mathbb{R}$$ ?

/Frank

 

[daveb]

So what is your question? What is puzzling you?

 

[LeonhardEuler]

By the test of comparison

$$\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | < \frac{1}{n^2}$$

y can be 0, so this should be
$$\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}$$

 

[Curious3141]

Hello

I have this question here which has puzzled me.

Given a series

$$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}$$

Show that the series converge for every $$y \in \mathbb{R}$$

By the test of comparison

$$\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | < \frac{1}{n^2}$$

Since its know that

$$\sum \limit_{n=0} ^{\infty} \frac{1}{n^2}$$ converge, then the series $$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}$$ converge for every $$y \in \mathbb{R}$$

There's a problem - the sequence you're comparing with is zeta(2), which converges (the sum is pi^2/6). However, that sum goes from n = 1 to infinity. When you start with n = 0, the first term is infinite, so obviously the sequence diverges.

I would suggest amending it like so :

$$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} < \frac{1}{y^2} + \sum \limit_{n=1} ^{\infty} \frac{1}{n^2} = \frac{1}{y^2} + \frac{\pi^2}{6}$$

BTW, y must be nonzero for convergence.

 

[Math_Frank]

So What can I conclude that the series

doesn't converge for every $$y \in \mathbb{R}$$ ??

/Frank

There's a problem - the sequence you're comparing with is zeta(2), which converges (the sum is pi^2/6). However, that sum goes from n = 1 to infinity. When you start with n = 0, the first term is infinite, so obviously the sequence diverges.

I would suggest amending it like so :

$$\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} < \frac{1}{y^2} + \sum \limit_{n=1} ^{\infty} \frac{1}{n^2} = \frac{1}{y^2} + \frac{\pi^2}{6}$$

BTW, y must be nonzero for convergence.

 

[Curious3141]

So What can I conclude that the series

doesn't converge for every $$y \in \mathbb{R}$$ ??

/Frank

It converges for all $$y \in \mathbb{R}$$\{0}

 

[Math_Frank]

It converges for all $$y \in \mathbb{R}$$\{0}

Okay,

what about my other conclusions regarding the series ?
Do they look okay?

/Frank

 

[Curious3141]

Okay,

what about my other conclusions regarding the series ?
Do they look okay?

/Frank

I honestly do not know enough analysis to comment on those, I'm sure someone more knowledgeable will be along to handle those.

 

[benorin]

For all real $$y\neq 0,$$ we have

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{\pi}{2y} \mbox{coth}( \pi y)$$​

EDIT: my bad, I forgot to start with n=0, but hey: who's counting.

 

[Math_Frank]

Hello benorin,

Could You please check if my other statements about Uniform Convergence and continuety for the series, are correct?

/Frank

For all real $$y\neq 0,$$ we have

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{\pi}{2y} \mbox{coth}( \pi y)$$​

EDIT: my bad, I forgot to start with n=0, but hey: who's counting.

 

[benorin]

Convergence is uniform for all non-zero real y, since

$$\left| \frac{1}{y^2 + n^2}\right| \leq \frac{1}{n^2}$$ for all $$n\geq 1$$

and $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ converges, thus

EDIT: forgot a + sign in this one:

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{1}{y^2}+ \sum \limit_{n=1} ^{\infty} \frac{1}{y^2 + n^2}$$

converges uniformly for all $$y\neq 0$$ EDIT: by the Weierstrass M-test. Furthermore, as each term in the series is a continuous function of y, the sum function is also continuous by uniform convergence.

Note that you can derive the formula

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} =\frac{\pi}{2y} \mbox{coth}( \pi y)$$​

from the partial fraction expansion of coth(x), which is

$$\mbox{coth}( x)= 2x\sum \limit_{n=1} ^{\infty} \frac{1}{x^2 + \pi ^2n^2}+\frac{1}{x}$$​

 

[Math_Frank]

Okay Thank You for correcting me :)

One final question though

I need to show that

$$lim _{y \rightarrow \infty} f(y) = 0$$

Can I Do this by Wierstrass M-Test?

/Frank

Convergence is uniform for all non-zero real y, since

$$\left| \frac{1}{y^2 + n^2}\right| \leq \frac{1}{n^2}$$ for all $$n\geq 1$$

and $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ converges, thus

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{1}{y^2} \sum \limit_{n=1} ^{\infty} \frac{1}{y^2 + n^2}$$

converges uniformly for all $$y\neq 0$$. Furthermore, as each term in the series is a continuous function of y, the sum function is also continuous by uniform convergence.

Note that you can derive the formula

$$f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} =\frac{\pi}{2y} \mbox{coth}( \pi y)$$​

from the partial fraction expansion of coth(x), which is

$$\mbox{coth}( x)= 2x\sum \limit_{n=1} ^{\infty} \frac{1}{x^2 + \pi ^2n^2}+\frac{1}{x}$$​

 

[benorin]

Use uniform convergence, the main idea of uniform convergence is to allow the interchange of limiting processes, such as

$$\lim_{x\rightarrow \infty}\sum_{n=0}^{\infty}f_n(x) = \sum_{n=0}^{\infty}\lim_{x\rightarrow \infty}f_n(x)$$

if the series is uniformly convergent for all x. Try it.

 

[Math_Frank]

Then by Uniform Convergence

$${\lim \sup_{n \rightarrow \infty}} \left| f_n(y) - f(y) \right| = 0$$

Can I then conclude that

$${\lim_{n \rightarrow \infty}}f(y) = 0$$ ??

Use uniform convergence, the main idea of uniform convergence is to allow the interchange of limiting processes, such as

$$\lim_{x\rightarrow \infty}\sum_{n=0}^{\infty}f_n(x) = \sum_{n=0}^{\infty}\lim_{x\rightarrow \infty}f_n(x)$$

if the series is uniformly convergent for all x. Try it.

 

[benorin]

What is meant is, do this:

$$\lim_{y\rightarrow \infty}f(y)=\lim_{y\rightarrow \infty}\sum_{n=0}^{\infty} \frac{1}{y^2 + n^2}= \sum_{n=0}^{\infty}\lim_{y\rightarrow \infty} \frac{1}{y^2 + n^2} = \sum_{n=0}^{\infty}0=0$$​