Uniform Convergence of $\sum_{k=0}^\infty\frac{kx}{1+k^4x^2}$

Question

This problem arose in trying to establish that $$f(x)=\sum_{k=0}^\infty\frac{kx}{1+k^4x^2}$$ is uniformly convergent on $[a,\infty), a>0$, and I thought I could do this, as I was able to show that since $\displaystyle\sup_{x\in\mathbb{R}}\left\{\displaystyle\frac{kx}{1+k^4x^2}\right\}=\displaystyle\frac{1}{2k}$ then this is uniform convergent if:

$$\lim_{n\to\infty}\sum_{k=n+1}^\infty\frac{1}{2k} =0$$

But the above limit doesn't make sense to me ... is it even defined? - I suspected not, so where should I go from here?

Answer 1

This limit isn't defined, and the series doesn't converge uniformly on $\mathbb R$.

On $[a, \infty)$, though, (assuming $k > 0$) we have $\frac{kx}{1 + k^4 x^2} = \frac{k}{1 / x + k^4 x} < \frac{1}{k^3 a}$, and series $\sum\limits_{k=1}^\infty \frac{1}{k^3 a}$ converges.

Answer 2

Note that $f'(x)=\frac{k(1-k^4x^2)}{1+k^4x^2}$. So $f$ is strictly decreasing for $x^2\leq \frac{1}{k^4}$. Hence, for all $k\geq \frac{1}{a^{1/4}},\,\sup_{x\in [a,\infty)}\frac{kx}{1+k^4x^2}=\frac{ka}{1+k^4a^2}$. Now $\sum_k\frac{ka}{1+k^4a^2}$ is convergent, so $\sum_k\frac{kx}{1+k^4x^2}$ is uniformly convergent on $[a,\infty)$ for $a>0$.

Answer 3

This is an extended comment on the sum

$$f(x) = \sum_{k=0}^\infty\frac{kx}{1+k^4x^2}$$

Result

It might be interesting that there is a closed expression for the sum which is given by

$$f(x) = \frac{1}{4} i \left(\psi ^{(0)}\left(1+\frac{\sqrt[4]{-1}}{\sqrt{x}}\right)-\psi ^{(0)}\left(1+\frac{(-1)^{3/4}}{\sqrt{x}}\right)\\+\psi ^{(0)}\left(1-\frac{\sqrt[4]{-1}}{\sqrt{x}}\right)-\psi ^{(0)}\left(1-\frac{(-1)^{3/4}}{\sqrt{x}}\right)\right)\tag{1}$$

where $\psi ^{(0)}\left(z\right)$ is the polygamma function.

This can also be written using the harmonic number function $H_z$ for complex $z$ as

$$f(x) = -\frac{1}{2} \Im\left(H_{+\left(\sqrt[4]{-1} /\sqrt{x}\right)}+H_{-\left(\sqrt[4]{-1} /\sqrt{x}\right)}\right)\tag{2}$$

The limits at $x=0$ are, as conjectured in my comment,

$$\lim_{x\to \pm 0}f(x) = \pm \frac{\pi}{4}\tag{3}$$

Here's a graph of the function

Remark: In the meantime I have verified (e.g. by expanding the sum around $x=1$) that the little bump close to $|x| \simeq 1$ is correct. In fact $f'(0.45) \simeq 0$ so that there is a local maxium. The second derivative has a pronounced negative minimum of $-0.335$ at $x \simeq 0.7$.

For values from $b=5$ (see discussion) on there is a more and more pronounced maximum in the $f(x)$. Hence the little bump is just an "announcement" of bigger bumps to come.

Derivation

Considering the partial sum we can use partial fraction decomposition to write

$$f(x,m)=\sum_{k=0}^{m}\frac{kx}{1+k^4x^2}\\ =\frac{1}{2} \sum_{k=0}^{m}\frac{kx}{1- i x k^2 }- \frac{1}{2} \sum_{k=0}^{m}\frac{kx}{1+ i x k^2 } $$

Repeating partial fraction decomposition with these terms leads finally to four sums of the type

$$\sum _{k=1}^m \frac{k x}{k \sqrt{i x}-i} = -i m \sqrt{i x}+\psi ^{(0)}\left(m-\frac{\sqrt{i x}}{x}+1\right)-\psi ^{(0)}\left(1-\frac{\sqrt{i x}}{x}\right)$$

Diverging terms cancel, and the limit of the partial sum $\lim_{m\to\infty} f(x,m)$ can be done employing the asymptotic expansion of $\psi $ with the result given in $(1)$.

Discussion

1. $x \to \infty$

The asymptotic behaviour for large $x$ is

$$f(x)_{x\to\infty} \simeq \zeta(3)/x - \zeta(7)/x^3 + O(1/x^5)$$

This can be obtained immediately from the sum representation.

2. Derivatives at $x=0$

From the analytic expression we can derive the following power series expansion abount $x=0$

$$f(x) = \frac{\pi}{4} \text{sign}(x) -\frac{x}{12} + \frac{x^3}{3!\; 42} - \frac{10 x^5}{5!11 } \pm ...\tag{d1}$$

For $x>0$ the derivative of the sum is

$$f'(x) = \sum_{k=0}^\infty\frac{k}{1+k^4x^2}- \sum_{k=0}^\infty\frac{k x k^4 2 x}{(1+k^4x^2)^2}\tag{d2}$$

If we let $x\to 0$ we obtain from comparng $(d1)$ and $(d2)$ formally the equation

$$f'(0) = \sum_{k=0}^\infty\ k = -\frac{1}{12}\tag{d3}$$

This is the famous Ramanujan relation which can be obtained by analytic continuation of the Riemann zeta function $\zeta(s) =\sum_{k=1}^\infty k^{-s}$ to $s=-1$: $\zeta(-1) = -\frac{1}{12}$.

Similar strange relations are obtained from the higher derivatives

$$\sum k^5 \simeq - \frac{1}{3! \;42}, \sum k^9 \simeq - \frac{10}{5! \;11}, ...$$

3. Similar sums

Consider sums with a qualitatevely similar appearance of the form

$$f_{a,b}(x) = \sum_{k=1}^\infty \frac{x k^a}{1+ x^2 k^b}$$

This first example is simpler has a simpler closed expression

$$f_{0,2}(x) = \sum_{k=1}^\infty \frac{x}{1+ x^2 k^2} = \frac{1}{2} \left(\pi \coth \left(\frac{\pi }{x}\right)-x\right)$$

We have

$$\lim_{x \to \pm 0} f_{0,2}(x) = \pm \frac{\pi}{2}$$

and

$$f_{0,2}(x)_{x\to\infty} \simeq \zeta(2)/x - \zeta(4)/x^3 + O(1/x^5)$$

It seems that the sum

$$f_{1,3}(x) = \sum_{k=1}^\infty \frac{x k}{1+ x^2 k^3}$$

is a more complicated case. But by using the heuristic trick of replacing the sum by an integral we obtain the expression

$$\int_0^{\infty } \frac{k x}{k^3 x^2+1} \, dk = \frac{2 \pi }{3 \sqrt{3} x^{\frac{1}{3}}}$$

which is the conjectured behaviour close to $x=0$.

Remark: the "integral trick" was correct also for the other two example sums and it is not a trick but can be justification strictly using the Euler-Maclaurin formula.

Answer 4

More generally if there exists sequences $\epsilon_0 >0, x\in D, $and $n_k, x_{n_k}$ such that x_{n_k} converges to x but $|f_{n_k}(x_{n_k})-f(x)| >\epsilon_0$ then $f_n(x_n)$ doesn’t converge uniformly to f