# Limit on a certain double sum

While working with multi-zeta functions, I encountered the below (experimental) value for a certain evaluation (in a limit sense). Notice first this well-known fact in context $$\sum_{n,m\geq1}\frac1{nm(n+m)}=2\zeta(3).$$

QUESTION. Does this hold true? If yes, how? $$\lim_{s\rightarrow\frac12^+}\sum_{n,m\geq1}\frac{2s-1}{n^sm^s(n+m)}=\pi.$$

Edited. $$s\rightarrow\frac12$$ has been replaced by $$s\rightarrow\frac12^+$$.

• Did you try expressing that sum in terms of the polylogarithm function (and look for a convenient known identity)? Jul 27 at 18:40
• I did but with no convincing progress. Of course, you might luck out. Jul 27 at 18:43

The limit (as $$s\downarrow1/2$$) will not change if the double sum is replaced by the corresponding double integral, for which we find this, as desired:

Alternatively, the limit (as $$s\downarrow1/2$$) will not change if the double sum is replaced by the double integral $$I(s):=(2s-1) \iint\limits_{x,y>0,\,x+y>1}\frac{dx\,dy}{x^s y^s(x+y)} \\ =(2s-1)\int_1^\infty \frac{du}u\,\int_0^u \frac{dx}{x^s (u-x)^s} \\ =(2s-1)\int_1^\infty \frac{du}{u^{2s}}\,J(s) =J(s),$$ where $$J(s):=\int_0^1 \frac{dt}{t^s (1-t)^s}\to J(1/2)=\pi.$$ So, $$I(s)\to\pi$$, as desired.

Details on why the limit (as $$s\downarrow1/2$$) will not change if the double sum is replaced by an appropriate double integral: The reasons for this are as follows:

1. $$f(x,y):=\dfrac1{x^s y^s(x+y)}$$ is decreasing in $$x>0$$ for each real $$y>0$$ and in $$y>0$$ for each real $$x>0$$. So, for any natural $$i$$ and $$j$$, $$f(i,j)\ge\int_{(i,i+1]\times(j,j+1]}f\ge f(i+1,j+1)$$.

2. For each natural $$m$$, $$\sum_{n\ge1}f(m,n)<\infty$$ and, similarly, for each natural $$n$$, $$\sum_{m\ge1}f(m,n)<\infty$$. Each of such ordinary sums will get multiplied by $$2s-1$$, and so, will not contribute to the limit as $$s\downarrow1/2$$.

2a. Similarly to point 2, for each real $$x>0$$, $$\int_{\max(0,1-x)}^\infty f(x,y)\,dy\le\dfrac1{x^s}\int_0^\infty \min\Big(\dfrac1{y^s},\dfrac1{y^{s+1}}\Big)\,dy =\dfrac1{(1-s)sx^s}$$, and $$\int_{\max(0,1-y)}^\infty f(x,y)\,dx\le\dfrac1{(1-s)sy^s}$$ for each real $$y>0$$.

• I'm likely missing something obvious, but why is it necessarily so that the double sum and double integral have the same limit? Jul 27 at 21:06
• @StevenStadnicki : I have now added details on that. Jul 27 at 21:48