Properties of the function defined by $g(x) = \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2}$

Question

I am looking at the function $g:\mathbb{R} \rightarrow \mathbb{R}$ defined as $$g(x) = \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2}$$

I would like to know if this function is convergent, continuous and differentiable.

For convergence I have $g(0)$ is divergent because we get a sum of 1, but for $x \not = 0$ we get $$0< \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2} < \sum\limits_{n=0}^{\infty} \frac{1}{n^2x^2} = \frac{1}{x^2} \sum\limits_{n=0}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6x^2}$$ and since $g$ is monotonic and bounded it converges (right?)

Since the pointwise limit doesn't exist for $x \not = 0$ then $g$ doesn't converge uniformly making proving continuity more difficult.

For differentiability I know that if $g_m(x)$ (partial sum of g) is point-wise convergent and $g_m'(x)$ is uniformly convergent then $g'(x) = lim_{m \rightarrow \infty } g'_m(x)$ but again I can't use this due to pointwise limit not existing.

For any of these properties could I look at $g(x)|_{(0,\infty)}$ and $g(x)|_{(-\infty,0)}$ and make conclusions from this?

Any guidance or solutions would be helpful,

Thanks.

Using Zeta function regularization you can define $g(0)$ to be $-1/2$. — Ali Caglayan, Apr 27 '14 at 18:02
Also you did mean for the index in the sums to be $n$ right? — Ali Caglayan, Apr 27 '14 at 18:06
Thanks for your reply. I got this question from an old exam paper (that I am sitting tomorrow) and since Zeta function regularization isn't in the module I would prefer to avoid using it. It should be solvable either by definitions results about uniform convergence of partial sums implying continuity (if the partial sums are continuous) and the differentiability result as put in my question — , Apr 27 '14 at 18:08
possible duplicate of Differentiability of $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{1+xn^{2}}\right)$ on $(0,+\infty)$ — Andrés E. Caicedo, Apr 27 '14 at 18:16

score 2 · Accepted Answer · answered Apr 27 '14 at 18:10

2

First notice that $g$ is even so it suffices to studied it on the interval $[0,\infty)$. Clearly $g$ isn't defined on $0$ and for $x>0$ we have $$g_n(x)=\frac1{1+n^2x^2}\sim_\infty\frac{1}{x^2}\frac1{n^2}$$ and since the series $$\sum_{n\ge1}\frac1{n^2}$$ is a convergent Riemann series then $g$ is defined on $(0,\infty)$.

Now for the uniform convergence, let $a>0$ then $$\frac1{1+n^2x^2}\le\frac1{1+n^2a^2},\quad\forall x\ge a$$ then we have the uniform convergence on every interval $[a,+\infty)$ and so the function $g$ is continuous on $(0,+\infty)$. There's not a uniform convergence on $(0,+\infty)$ since $$\lim_{x\to0}g(x)$$ doesn't exist. We can prove that $g$ is $\mathscr C^\infty$ on $(0,+\infty)$ since we can prove that the series $$\sum_{n\ge0}g_n^{(k)}(x),\quad \forall k\ge1$$ is uniformly convergent on every compact in $(0,\infty)$.

answered Apr 27 '14 at 18:10

For uniform convergence on $[a,\infty]$ is this a result from the M-test? and then this proves continuity on on $[a,\infty]$ for $a>0$ so its continuous on $(0,\infty)$? What do you mean by every compact in $(0,\infty)$? Thanks for your reply – Apr 27 '14 at 18:18
Yes this is from the M-test and since the and since the continuity is a local notion then we have the continuity on $(0,\infty)$. And for you last question: I mean every compact $[a,b]\subset (0,\infty)$. – Apr 27 '14 at 18:22
I understand, thanks for your help! – Apr 27 '14 at 18:31
You're welcome. – Apr 27 '14 at 18:32

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

If $|x|\gt r$, we have that $$ \begin{align} \left|\,\sum_{n=N}^\infty\frac1{1+n^2x^2}\,\right| &\le\frac1{r^2}\sum_{n=N}^\infty\frac1{n^2}\\ &\le\frac1{r^2}\sum_{n=N}^\infty\frac1{n(n-1)}\\ &=\frac1{r^2(N-1)} \end{align} $$ so the sum converges uniformly outside of any neighborhood of $0$.

Furthermore, using this answer, we get $$ \begin{align} \sum_{n=0}^\infty\frac1{1+n^2x^2} &=\frac1{x^2}\sum_{n=0}^\infty\frac1{\frac1{x^2}+n^2}\\ &=\frac1{2ix}\sum_{n=0}^\infty\left(\frac1{n-\frac ix}-\frac1{n+\frac ix}\right)\\ &=\frac12-\frac1{2ix}\sum_{n\in\mathbb{Z}}\frac1{n+\frac ix}\\ &=\frac12-\frac1{2ix}\pi\cot\left(\pi\frac ix\right)\\ &=\frac12+\frac\pi{2x}\coth\left(\frac\pi{x}\right) \end{align} $$

Properties of the function defined by $g(x) = \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2}$

2 Answers2