$\newcommand{\d}{\,\mathrm{d}}$I've done my best with this series, but I've never actually seen a sum of square roots before!
The following question was given to me by a friend - apparently some author of theirs had left the release date of some event in the answer. I pursued it not for that, but rather because there is an interesting lesson (for me) in series evaluation to be learnt here - I give the above context only to explain how I know the answer is $2+22i$, and I know that the answer is possible to get (albeit maybe only empirically).
Let: $$\Gamma:=\left\{f_n:\Bbb C\to\Bbb C,\,z\mapsto\frac{\pi^2}{24}z^2-\frac{1}{n^2}z+\frac{589}{20}\cdot\frac{1}{n^3}\,\Big|\,n\in\Bbb N_1\right\}$$Let $\Lambda$ be the set of all roots, with positive imaginary part, of the elements of $\Gamma$. Find the sum of all elements of $\Lambda$, rounding the real and imaginary parts to the nearest integer.
The quadratic formula easily yields that each root $\lambda_n\in\Lambda$, corresponding to $f_n\in\Gamma$, is:
$$\lambda_n=\frac{12}{\pi^2}\cdot\frac{1}{n^2}\left(1+\sqrt{1-\pi^2\cdot\frac{589}{120}n}\right)$$
Where any branch of the square root is used which maps the negative real axis to the positive imaginary one. I think the branch choice will become more important in summation later.
Then the desired answer is:
$$\sum_{\lambda\in\Lambda}\lambda=\sum_{n=1}^\infty\lambda_n=\frac{12}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2}+i\cdot\frac{12}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2}\sqrt{\pi^2\cdot\frac{589}{120}n-1}=2+i\cdot\frac{12}{\pi^2}S$$
Where $S=\sum_{n=1}^\infty\frac{1}{n^2}\sqrt{\alpha n-1}$ is the heart of the problem, for some real $\alpha\gt1$ (here of course $\alpha=\pi^2\cdot\frac{589}{120}$ but the specific detail is irrelevant I think). Empirically one quickly sees that, after rounding, $\frac{12}{\pi^2}S\approx22$, but I wonder if there is an analytic solution.
It is not possible to recast this as a Riemann sum; the only things I think have a chance of working are contour integration, or Fubini-ing a double series of positive terms where I used the square root's binomial expansion.
I'll show the latter first:
$$|z|\lt1:\sqrt{1-z}=1-\sum_{m=1}^\infty\binom{2m-2}{m-1}\frac{1}{2^{2m-1}m}z^m$$
Re-arrange $S$ to obtain:
$$S=\pi\sqrt{\frac{589}{120}}\sum_{n=1}^\infty\frac{1}{n^2\sqrt{n}}\sqrt{1-\frac{1}{\pi^2}\frac{120}{589n}}=\pi\sqrt{\frac{589}{120}}S'$$
And then:
$$\begin{align}S'&=\sum_{n=1}^\infty\frac{1}{n^2\sqrt{n}}-\sum_{n=1}^\infty\frac{1}{n^2\sqrt{n}}\sum_{m=1}^\infty\binom{2m-2}{m-1}\frac{1}{2^{2m-1}m}\frac{1}{(\alpha n)^m}\\&=\zeta\left(\frac{5}{2}\right)-\sum_{m=1}^\infty\binom{2m-2}{m-1}\frac{1}{2^{2m-1}m}\frac{1}{\alpha^m}\zeta\left(\frac{5}{2}+m\right)\end{align}$$
But this is not any power series that I recognise. It is perhaps possible to split it as a Cauchy product but I can't spot any such pattern.
Another approach is to consider contour integration with the ever useful $\pi\cot\pi z$: let then $f(z)=\sqrt{\alpha z-1}z^{-2}$. Unfortunately, there is a branching issue. Not being very experienced with such issues, I propose maybe that one should take a branch cut of the square root along the negative imaginary axis, i.e. $$\sqrt{\cdot}:z\mapsto\begin{cases}\exp\left(\frac{1}{2}\log z\right)&\text{otherwise}\\\exp\left(\frac{1}{2}\log z+\pi i\right)&\Im z\lt0\wedge\Re z\le0\end{cases}$$
For $\log$ the principal branch.
We cannot do the usual big box method here, but I think one can adapt it slightly. Let the contour $\gamma_{\epsilon,N}$, for natural $N$ and real $\epsilon\gt0$, extend first from $\epsilon-\frac{1}{2}i\to(N+1/2)-\frac{1}{2}i$ along the positive real axis, then vertically upward to $(N+1/2)+(N+1/2)i$, then horizontally go east to $-(N+1/2)+(N+1/2)i$, then down to $-(N+1/2)-\frac{1}{2}i$, then horizontally right to $-\epsilon-\frac{1}{2}i$. Lastly, some keyhole contour that goes up from $-\epsilon-\frac{1}{2}i$, wrapping around $0$ with constant radius $\epsilon$, then down again to $\epsilon-\frac{1}{2}i$. I choose this because it avoids the branch cut, avoids the poles of $f$ and of $\cot$, yet it includes them (except the one at $0$) for the sake of residue summation and most of the contour disappears for large $N$. Letting the keyhole part be denoted $\gamma'$, we get:
$$\begin{align}\lim_{N\to\infty}\oint_{\gamma_{N,\epsilon}}f(z)\cdot\pi\cot\pi z\d z&=2\pi i\sum_{n\in\Bbb Z\setminus\{0\}}\frac{1}{n^2}\sqrt{\alpha n-1}\\&=\int_{-\infty-i/2}^{-\epsilon-i/2}\pi\cot\pi x\cdot f(x)\d x+\int_{\epsilon-i/2}^{\infty-i/2}\pi\cot\pi x\cdot f(x)\d x+\oint_{\gamma'_{\epsilon}}\pi\cot\pi z\cdot f(z)\d z\end{align}$$
Now I don't know much about more difficult contours, such as keyhole integration, as I have never been taught it, but I am hoping that by proximity with $0$ the cotangent goes very close to $0$ and the integral vanishes as $\epsilon\to0$. Not very mathematical I know, but I'm out of my depth here! Assuming this is vaguely correct, noting then for negative integers the residue summand is imaginary, and real otherwise, one finds:
$$\begin{align}S&=\frac{1}{2}\Im\left[\int_{-\infty-i/2}^{\infty+i/2}\cot\pi z\cdot f(z)\d z\right]\end{align}$$
But I quickly realise that I have no hope of evaluating this integral, even after a shift onto the real axis, since after a separation into real and imaginary parts one is left with an ugly mess.
How can you approach series of the form $\sum\frac{1}{n^k}\sqrt{\alpha n-1}$? Was either approach of mine salvageable?
But why not use the integral approximation
$$S\approx\int_{1}^{\infty}\frac{\sqrt{\alpha x-1}}{x^{2}} $$
Or after computing the integral
$$S\approx\frac{\alpha\pi}{2}-\alpha\arctan{\sqrt{\alpha-1}}+\sqrt{\alpha -1}$$