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Let $$f(n) = 2 \sum\limits_{a = 2}^{n - 1} \sum\limits_{b = n + 1}^{n + a - 1}\frac{1}{ab} .$$

One can see that $$\lim\limits_{n \to \infty}f(n) = 2\int\limits_0^1 \frac{dx}{x} \int\limits_1^{1 + x}\frac{dy}{y} = \frac{\pi^2}{6}.$$ This suggests that the double sum defining $f(n)$ can be transformed into a single sum resembling the inverse square sum (although I didn't quite manage to do it after a few minutes with pencil and paper).

Question: Is there a simple closed form for $f(n)$ so that the limit value (or existence, for that matter) is evident?

Motivation: consider a random "wannabe-graceful-tree" graph $G = (V, E)$ with $V = [n]$, and $E$ containing a single random pair $\{x, x + d\}$ for each $d$ from $1 \to n - 1$. Then $f(n)$ is (almost) the expectation of the number of triangles in $G$.

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    $\begingroup$ $$f(n) = 2\int_0^1 dy \frac{y^{n-1}}{1-y} \int_y^1 dx\frac{x-x^{n-1}}{1-x}~~~ -~~~ 2\frac{H_{n-1}-1}{n}$$ $$ = 2\int_0^1 dx \frac{x-x^{n-1}}{1-x} \int_0^x dy \frac{y^{n-1}}{1-y}~~~ -~~~ 2\frac{H_{n-1}-1}{n}.$$ $\endgroup$
    – Max Alekseyev
    Commented Jun 28, 2018 at 13:39

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We have $$f(n+1)-f(n)=\frac1{n^2}+\frac2{n}\left(1+\frac 12+\ldots+\frac 1{n-1}\right)-\frac2{n+1}\left(1+\frac 12+\ldots+\frac 1{n}\right),$$ and $f(2)=0$. So \begin{gather*} f(m)=\sum_{n=2}^{m-1}(f(n+1)-f(n))=\sum_{n=2}^{m-1}\frac1{n^2}-\frac2{m}\left(1+\frac 12+\ldots+\frac 1{m-1}\right)+1=\\=\sum_{n=1}^{m-1}\frac1{n^2}-\frac2{m}\left(1+\frac 12+\ldots+\frac 1{m-1}\right). \end{gather*}

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  • $\begingroup$ Nice!! It might be useful to mention the generating function:$$\sum_nf(n)t^{n-1}=\frac{\operatorname{Li}_2(t)}{1-t}-\frac{\log^2(1-t)}t$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 1, 2018 at 12:15

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