# Math problem (analysis)

broegger
Hi,

I have to show that the function

$$f(x) = \sum_{n=1}^{\infty}\frac1{x^2+n^2}$$​

tends to 0 as $$x \rightarrow \infty$$, i.e. $$\lim_{x\rightarrow\infty}f(x) = 0$$. How can I do this?

There is a hint that says I should use the inequality $$f(x) \leq \sum_{n=1}^N\tfrac1{x^2+n^2} + \sum_{n=N+1}^\infty\tfrac1{n^2}$$. It is obvious that the first term approaches 0 as $$x \rightarrow \infty$$, but what about the second term?

## Answers and Replies

Homework Helper
How do you know that?

How do you know that

$$\lim_{x\rightarrow +\infty}\sum_{n=1}^{N}\frac{1}{x^{2}+n^{2}}=? 0$$

My Maple says it's undefined,as well as the initial limit...

Daniel.

Homework Helper
Denote the initial limit by $F =:\lim_{x\rightarrow +\infty} f(x)$

U can show that $f(0)=\zeta\left(2\right)=\frac{\pi^{2}}{6}$ and $\forall x\in\mathbb{R}, f(x)\leq f(0)$.

Daniel.

Homework Helper
I've looked at the graph and the function quicky goes to 0 as "x" increases.

It's continuous everywhere on $\mathbb{R}$ and has the derivative

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

which is negative on the positive semiaxis.Therefore the function decreases continuously on the positive semiaxis.Since it's continuous,even and strictly positive,the asymptotic limits have to be zero.

Looking forward for critique.

Daniel.

Staff Emeritus
Gold Member
but what about the second term?

It's a function of N.

(Notice that f(x) is not a function of N...)

Staff Emeritus
Gold Member
Therefore the function decreases continuously on the positive semiaxis.Since it's continuous,even and strictly positive,the asymptotic limits have to be zero.

Why wouldn't the asymptotic limit be 1/10, or &pi;4/792?

Homework Helper
You know what, I would give a whole dollar if someone could explain to me why:

$$\sum_{n=1}^{\infty}\frac{1}{a^2+n^2}=\frac{a\pi Coth[a\pi]-1}{2a^2}$$

This is what Mathematica returns and I don't wish to suggest using Mathemtaica to prove things but I don't know how to prove the original question, and in my attempts to learn how, I've reduced it to the above. Now, you know what happens to the RHS as a goes to infinity right?

Homework Helper
Interesting.My ancient version of Maple returned a difference of psi functions of complex arguments multiplied with 1/(ix).

Daniel.

broegger
Hurkyl said:
It's a function of N.

(Notice that f(x) is not a function of N...)

I can let N -> infinity and thereby make the second term vanish but then what about the first term which also depends on N?

Staff Emeritus
Gold Member
It's a finite sum.

Phoenix Assault
Hey in response to your question, I have observed that it's very easy to tell that the sum tends to 0 as n goes to infinity. As n^2 gets large, the x^2 term gets very large too at the same rate. Any # with an inreasing denominator decreases in value, therefore the total sum goes to zero. :rofl:

Last edited:
Staff Emeritus