# Some numerically divergent sum

I try to get a numerically evaluation of the following sum:

$$\sum _{j=0}^{\infty } \frac{j^2}{\sqrt{j \left(j^2+\frac{1}{2}\right)}}$$

using different methods. With Euler-Maclaurin we get

NSum[
j^2/Sqrt[j (j^2 + 1/2)]
, {j, 1, \[Infinity]}
, Method -> "EulerMaclaurin"
, WorkingPrecision ->  20]


with Levin u-transform we get the same result: Infinity

I think the correct sum is $$-0.78403507353706181$$ maybe it is part of some analytic continuation?

Start by noticing that the $$j=0$$ term is zero,

Limit[j^2/Sqrt[j (1/2 + j^2)], j -> 0]
(*    0    *)


so we can sum from $$j=1$$. Then extract the dominant singular term $$\sqrt{j}$$ from the sum,

Asymptotic[j^2/Sqrt[j (1/2 + j^2)], j -> ∞]
(*    Sqrt[j]    *)


for easier numerical convergence through analytic continuation:

$$\sum_{j=1}^{\infty}\frac{j^2}{\sqrt{j\left(j^2+\frac12\right)}} = \sum_{j=1}^{\infty}\sqrt{j} + \sum_{j=1}^{\infty}\left(\frac{j^2}{\sqrt{j\left(j^2+\frac12\right)}}-\sqrt{j}\right)\\ = \zeta\left(-\frac12\right) + \sum_{j=1}^{\infty}\left(\frac{j^2}{\sqrt{j\left(j^2+\frac12\right)}}-\sqrt{j}\right)$$

Zeta[-1/2] + NSum[j^2/Sqrt[j (1/2 + j^2)] - Sqrt[j],
{j, 1, ∞},
Method -> "EulerMaclaurin"] // Chop
(*    -0.783497    *)


By playing with NSum's parameters you may get more accurate values.

• It should be noticed that Sum[Sqrt[j], {j, 1, Infinity}, Regularization -> "Dirichlet"] performs Zeta[-(1/2)]. Aug 3, 2021 at 10:27