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}<br /> =\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}<br /> =\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​