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In the course of studying a certain complex-valued functional equation, I have had a need to evaluate the following limit:

$$\gamma_\mathcal{T}=\lim_{n\to\infty}\left(-\frac{i}{2}\sum_{k=1}^n \frac1{ik+k^{3/2}}-\log\left(1+\frac{i}{\sqrt n}\right)\right)$$

which is structurally similar to the usual limit definition for the Euler constant $\gamma$.

So far as I can tell, there seems to be no elementary closed form for this limit, so I set about trying for numerical estimation.

The problem is that the convergence of this limit looks to be excruciatingly slow. Even with the help of a sequence extrapolation method, I only managed to produce a few good digits:

$$\gamma_\mathcal{T}\approx-0.5-0.9300125396i$$

I am wondering if there are more efficient, alternative methods for numerically evaluating this limit. Thanks in advance!

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what is $i$, $\sqrt{-1}$? Then the logarithm tends to 0 and the series converges, right? –  Fedor Petrov Aug 6 '12 at 11:56

2 Answers 2

To expand on @Fedor's comment: if you rationalize the denominator, you get the general term of the sum to be $f(k)=\frac{k^{3/2} - i k}{k^3 + k^2}.$ The imaginary part of $\sum_{k=1}^\infty f(k)$ is easy to evaluate (it is equal to $-1$). The real part is not so easy, but Mathematica returns immediately with $1.8600250792211903071806\dots,$ so its sequence acceleration techniques are up to the task.

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2  
gp agrees: suminf(n=0,(-1)^n*(zeta(n+1.5)-1)) + .5 returns $1.8600250792211903071806959+$. –  Noam D. Elkies Aug 6 '12 at 18:27

As already mentioned by Fedor and Igor, you can ignore the logarithmic term since it zeroes out at $\infty$, and you can just concentrate on the series

$$-\frac{i}{2}\sum_{k=1}^\infty \frac1{ik+k^{3/2}}$$

Using Laplace transform techniques, your sum can be transformed into the integral

$$-\frac12-\frac{i}{\sqrt\pi}\int_0^\infty \frac{F(\sqrt{u})}{\exp\,u-1}\mathrm du$$

where $F(z)$ is Dawson's integral.

I don't know of a closed form for this integral, but Mathematica easily evaluates this numerically:

-1/2 - I NIntegrate[DawsonF[Sqrt[u]]/(E^u - 1), {u, 0, Infinity}, 
    Method -> "DoubleExponential", WorkingPrecision -> 50]/Sqrt[Pi]
-1/2 - 0.93001253961059515359034795785857166233326206076173 I
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