I was trying to get some interesting result for $\zeta(3)$, exploring the following function:
$$W(a) = \sum\_{k=1}^\infty \frac{1}{k^3 + a^3}, \mbox{ with } \lim\_{a\rightarrow 0} W(a) = \zeta(3).$$


Let $w\_1, w\_2, w\_3$ be the three roots (one real, two complex) of $(w+1)^3+a^3=0$, with $w\_1=-(a+1)$. Also, $a$ is a real number. Using Wolfram Alpha (see [here](https://www.wolframalpha.com/input/?i=sum%201%2F%28k%5E3%20%2Ba%5E3%29%20for%20k%3D1..infinity)), I get


$$W(a)=\frac{-1}{3}\cdot\sum\_{j=1}^3 W\_j(a), \mbox{ with } W\_j(a) = \frac{\psi^{(0)}(-w\_j)}{(w\_j+1)^2}.$$


Here $\psi^{(0)}$ is the digamma function. The result is wrong because $W\_1(a) \rightarrow \infty$ as $a\rightarrow 0^+$ while $W\_2(a)$ and $W\_3(a)$ remain bounded. Indeed using $a=0.0001$, Wolfram yields $W(a)\approx -2334.16$, see [here](https://www.wolframalpha.com/input/?i=sum%201%2F%28k%5E3%20%2B%280.0001%29%5E3%29%20for%20k%3D1..infinity). Surprising, with $a=0.01$ it yields $W(a)\approx 1.20206$ which is very close to the true result.


Surprisingly, Wolfram knows (see [here](https://www.wolframalpha.com/input/?i=lim%28sum%201%2F%28k%5E3%20%2B%20a%5E3%29%20for%20k%3D1..infinity%29%2C%20a%3D0)) that
$$\lim\_{a\rightarrow 0} W(a) = -\frac{\psi^{(2)}(1)}{2}.$$


Of course (this is a well known fact), $\zeta(3)=-\psi^{(2)}(1)/2$ and thus Wolfram is correct this time.


**My question:**


What is going on with this computation (or is it me?), and what is the correct formula for $W(a)$?


**Update**


See the two answers below proving that I was wrong, and that the Mathematica formula I though was incorrect, is indeed right. Kudos Mathematica! You were successful at solving a nice problem involving a few challenging steps, and coming with a somewhat unexpected but neat formula involving derivatives of the digamma function instead of the classic $\zeta(3)$.


**Final note**


It is possible to use a different, simpler approach that does not involve complex numbers. Consider


$$V(a) =\sum\_{k=1}^\infty (-1)^{k+1}\frac{1}{k(k^2-a^2)}.$$


Wolfram is able to compute the limit of $V(a)$ as $a\rightarrow 0$, and returns the correct value $3\zeta(3)/4$, see [here](https://www.wolframalpha.com/input/?i=lim%20%28sum%20%28-1%29%5E%28k%2B1%29%2F%28k%5E3-a%5E2%20k%29%2C%20k%3D1..infinity%29%2C%20a%20%3D%200). It is easy to establish that


$$V(a)=\frac{1}{a^2} \Big[\int\_0^\infty \frac{\cosh(ax)}{1+e^x} dx -\log 2\Big].$$


To compute $\lim\_{a\rightarrow 0} V(a)$, we apply L'Hospital Rule twice to the above expression, the denominator in this case being $a^2$. This yields


$$\lim\_{a\rightarrow 0}V(a) = \frac{1}{2}\lim\_{a\rightarrow 0}\int\_0^\infty \frac{x^2\cosh(ax)}{1+e^x}dx =\frac{1}{2}\int\_0^\infty \frac{x^2}{1+e^x}dx=\frac{3\zeta(3)}{4}.$$


Here, we assume $a<1$.



Erroneous Wolfram result for $\sum_{k=1}^\infty (k^3 + a^3)^{-1}$, looking for correct formula
I think the statement in the OP that $W\_2(a)$ and $W\_3(a)$ remain bounded when $a\rightarrow 0$ is mistaken, so that there is no inconsistency with the Mathematica result.


The three roots of $(w+1)^3+a^3=0$ are
$$w\_1= -a-1,\;\; w\_2= \tfrac{1}{2} \left(-i \sqrt{3} a+a-2\right),\;\;w\_3= \tfrac{1}{2} \left(i \sqrt{3} a+a-2\right).$$
Then the denominator $(w+1)^2$ vanishes for all three roots when $a\rightarrow 0$, while the numerator remains finite (equal to $-\gamma\_{\rm Euler}$).


And indeed, a numerical check suggets that the Mathematica output is actually correct, and the erroneous numerical result for small $a$ is a numerical instability in the computation of the digamma function. See these two plots that compare the digamma expression (blue) with a numerical evaluation of the sum (gold), as a function of $a$. For $a\gtrsim 0.01$ the two answers are nearly indistinguishable.


![](https://ilorentz.org/beenakker/MO/digammacomparison.png)
![](https://ilorentz.org/beenakker/MO/digamma_comparison_2.png)
