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How can I obtain an asymptotic for this partial sum, with an error term of at most $O(m^{1/2}\ln(m))$:

$$\sum_{n=1}^{m-1}\ln(n)\ln(m-n)$$

I tried flipping the order of summation half way through to rewrite it as: $\sum_{n=1}^{m-1}\ln(n)\ln(m-n)=O(\ln(m)^2)+2\sum_{n\leq m/2}\ln(n)\ln(m-n)$ $=O(\ln(m)^2)+2\ln(m)\sum_{n\leq m/2}\ln(n)+2\sum_{n\leq m/2}\ln(1-\frac{n}{m})\ln(n)$

So I could use a crude version of sterlings formula for the sum on the left and then just drop the sum on the right, but even this still only gives me an asymptotic with an error term of $O(m\ln(m))$.

Antonio Vargas
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Ethan Splaver
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1 Answers1

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First rewrite the sum as

$$ \begin{align} &\sum_{n=1}^{m-1} \ln(n)\ln(m-n) \\ &\qquad = \ln(m)\ln((m-1)!) + \sum_{n=1}^{m-1} \ln(n)\ln\left(1-\frac{n}{m}\right) \\ &\qquad = \ln(m)\ln((m-1)!) + \ln(m)\sum_{n=1}^{m-1}\ln\left(1-\frac{n}{m}\right) + \sum_{n=1}^{m-1}\ln\left(\frac{n}{m}\right)\ln\left(1-\frac{n}{m}\right) \\ &\qquad = 2\ln(m)\ln((m-1)!) - (m-1)\ln(m)^2 + \sum_{n=1}^{m-1}\ln\left(\frac{n}{m}\right)\ln\left(1-\frac{n}{m}\right). \end{align} $$

To estimate the new sum we can use the Euler-Maclaurin formula (as in this answer). This yields

$$ \sum_{n=1}^{m-1}\ln\left(\frac{n}{m}\right)\ln\left(1-\frac{n}{m}\right) = m\int_0^1 \ln(x)\ln(1-x)\,dx - \epsilon(m), $$

where

$$ \epsilon(m) = \int_0^1 \left(\lceil mx\rceil - mx - \frac{1}{2}\right)\left(\frac{\ln(1-x)}{x}-\frac{\ln(x)}{1-x}\right)\,dx = O(1). $$

Actually we should have $\epsilon(m) = o(1)$ as well but I haven't found a proof. Anyway, according to Mathematica

$$ \int_0^1 \ln(x)\ln(1-x)\,dx = 2 - \frac{\pi^2}{6}, $$

so this implies that

$$ \sum_{n=1}^{m-1} \ln(n)\ln(m-n) = 2\ln(m)\ln((m-1)!) - (m-1)\ln(m)^2 + \left(2-\frac{\pi^2}{6}\right)m + O(1). $$

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Antonio Vargas
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