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.
Best Answer
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 $-100.826666\ldots$.