The zeroes of the Harmonic numbers

Question

After I found that $$H_x=\sum_{k=1}^\infty\frac{x}{k(k+x)}$$I started to study its properties. I graphed this on desmos and found that it is zero for infinitely many values of $x$ is negative. How do I find when this sum is zero? Of course setting it equal to zero is one way to find out, but the only obvious solution is $x=0$.

Please give hints or approaches only. I really want to try this myself. Recurrence relations for the zeroes, or closed forms in terms of special functions (not defined by yourself), calculus symbols (derivatives, integrals, sums), asymptotics, and equations they satisfy other than $$\sum_{k=1}^\infty\frac{x}{k(k+x)}=0$$ are allowed.

This is not a duplicate to the question named "Find the Value of $\sum_{n=1}^\infty\frac a{n(n+a)}$" because it doesn't concern the zeroes of the function.

Edit: Why was I ignored? The above question is obviously not the same as mine. It might be concerning the same function but it has nothing to do with the zeroes of it. Take care to read the whole post before judging.

Answer 1

I shall derive a good asymptotics for $x_n$, the $n$th negative root. Since $$ \psi (1 + x) = \psi ( - x) - \pi \cot (\pi x), $$ and $\psi(-x) \sim \log(-x)$ for large negative $x$, $x_n$ satisfies $$\tag{1} \log ( - x_n ) - \pi \cot (\pi x_n ) \sim - \gamma . $$ We look for an expression of the form $$\tag{2} x_n = - n - \frac{1}{2} - f(n) $$ where $f(n)$ is bounded. Since $$ \log \!\Big( {n + \frac{1}{2} + f(n)} \Big) \sim \log\! \Big( {n + \frac{1}{2}} \Big) \sim H_n - \gamma $$ for large $n$, substitution of $(2)$ to $(1)$ yields $$ f(n) \sim \frac{1}{\pi }\arctan \left( {\frac{{H_n }}{\pi }} \right) $$ for large $n$. Hence $$\tag{3} x_n \sim - n - \frac{1}{2} - \frac{1}{\pi }\arctan \left( {\frac{{H_n }}{\pi }} \right) $$ for large $n$. As an example, $x_{100}= -100.826619\ldots$, whereas the approximation $(3)$ yields $-100.826666\ldots$.

Answer 2

Thanks to @prets, @JeanMarie, and @GregMartin's comments, I will put this partial answer. We know that $$H_x=\psi(x+1)+\gamma$$Where $\gamma$ is Euler's constant. Setting this expression equal to zero, we get $$\psi(x+1)=-\gamma$$ We know that $$\text{dom}(H_x)=\mathbb{R}/\mathbb{N}^-$$So we want to solve for the zeroes between the intervals $$-n-1<x<-n,n\in\mathbb{N}^+$$So we have that $$x=\psi^{-1}(-\gamma)-1$$Where $$\text{dom}(\psi^{-1}(x))=\{x|-(n+1)<x<-n,n\in\mathbb{N}^+\}$$

Answer 3

If we make $x=-t$, as already written in the post, answers and comments, for large $t$, we need to find the zero of $$\log (t)+\pi \cot (\pi t)+\gamma=0$$ In order to remove the discontinuities, it is better to consider that we look for the zeros of function $$f(t)=\gamma \sin (\pi t)+\pi \cos (\pi t)+\log (t) \sin (\pi t)$$ Expanded as a series around $t=n+\frac 12$ and using power series reversion $$\large\color{blue}{t_{(n)}=\left(n+\frac{1}{2}\right)+u-\frac{\pi ^2 (2 n+1)^2 \left(\log \left(n+\frac{1}{2}\right)+\gamma \right)+4}{2 (2 n+1) \left(\pi ^2 (2 n+1)-2\right)}\, u^2+O(u^3)}$$ where $$\color{blue}{u=\frac{\log \left(n+\frac{1}{2}\right)+\gamma }{\pi ^2-\frac{2}{2 n+1}}}$$

Some results

$$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 1.60050 & 1.60256 \\ 2 & 2.63815 & 2.64591 \\ 3 & 3.65643 & 3.67149 \\ 4 & 4.66607 & 4.68913 \\ 5 & 5.67096 & 5.70235 \\ 6 & 6.67295 & 6.71279 \\ 7 & 7.67305 & 7.72133 \\ 8 & 8.67183 & 8.72852 \\ 9 & 9.66968 & 9.73468 \\ 10 & 10.6668 & 10.7401 \\ \end{array} \right)$$

Edit

A better approximation can be obtained performing one single iteration of Halley or Householder method with $t_0=n+\frac 12$

$$t_1=t_0+\frac{2 t_0 \left(\pi ^2 t_0-1\right) (\log (t_0)+\gamma )}{\gamma ^2 \pi ^2 t_0^2+\log (t_0) \left(2 \gamma \pi ^2 t_0^2+\pi ^2 t_0^2 \log (t_0)+1\right)+2 \left(\pi ^2 t_0-1\right)^2+\gamma }$$

$$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 1.60084 & 1.60256 \\ 2 & 2.64031 & 2.64591 \\ 3 & 3.66172 & 3.67149 \\ 4 & 4.67536 & 4.68913 \\ 5 & 5.68486 & 5.70235 \\ 6 & 6.69185 & 6.71279 \\ 7 & 7.69721 & 7.72133 \\ 8 & 8.70143 & 8.72852 \\ 9 & 9.70483 & 9.73468 \\ 10 & 10.7076 & 10.7401 \\ \end{array} \right)$$

All of that is not as good as the superb answer from @Gary.

Answer 4

As I wrote in my other answer, with $t=-x$, we look for the zeros of function $$f(t)=\gamma \sin (\pi t)+\pi \cos (\pi t)+\log (t) \sin (\pi t)$$ Let $$t=\left(n+\frac{1}{2}\right)+\frac y \pi\quad \implies \quad g(y)=\cos (y) \left(\gamma+\log \left(n+\frac{1}{2}+\frac{y}{\pi }\right) \right)-\pi \sin (y)$$ Let Expanding for large values of $n$ $$g(y)=(\log (n)+\gamma ) \cos (y)- \pi \sinh(y)+O\left(\frac{1}{n}\right)$$ gives as a first approximation $$y_0=\tan ^{-1}\left(\frac{\log (n)+\gamma }{\pi }\right)$$ Using one iteration of Newton method gives $$y_1=y_0-\frac {g(y_0)}{g'(y_0)}$$ which is fully explicit.

Some numbers $$\left( \begin{array}{cccc} n & y_0 & y_1 & \text{solution} \\ 1 & 0.181707 & 0.323573 & 0.322215 \\ 2 & 0.384267 & 0.458623 & 0.458397 \\ 3 & 0.490035 & 0.538815 & 0.538746 \\ 4 & 0.558603 & 0.594190 & 0.594162 \\ 5 & 0.608059 & 0.635708 & 0.635694 \\ 6 & 0.646100 & 0.668498 & 0.668490 \\ 7 & 0.676650 & 0.695345 & 0.695341 \\ 8 & 0.701958 & 0.717916 & 0.717913 \\ 9 & 0.723417 & 0.737279 & 0.737278 \\ 10 & 0.741949 & 0.754160 & 0.754159 \\ \end{array} \right)$$