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I have to solve find the value of $\sum_{k=1}^{n/2} k\log k$ as a part of question.

How should I proceed on this ?

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    Just an idea if anybody could expand on it .... We can write k as n/2 in each of them .... then it would be that O(n/2 * log((n/2)factorial)) ..... Can anybody concise this expression more....2011-03-14

2 Answers 2

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Got it. The constant in Moron's answer is $C = \log A$, where $A$ is the Glaisher-Kinkelin constant. Thus C = \frac{1}{12} - \zeta'(-1).

The expression $H(n) = \prod_{k=1}^n k^k$ is called the hyperfactorial, and it has the known asymptotic expansion

$H(n) = A e^{-n^2/4} n^{n(n+1)/2+1/12} \left(1 + \frac{1}{720n^2} - \frac{1433}{7257600n^4} + \cdots \right).$ Taking logs and using the fact that $\log (1 + x) = O(x)$ yields an asymptotic expression for the OP's sum $\sum_{k=1}^n k \log k = C - \frac{n^2}{4} + \frac{n(n+1)}{2} \log n + \frac{\log n}{12} + O \left(\frac{1}{n^2}\right),$ the same as the one Aryabhata obtained with Euler-Maclaurin summation.


Added: Finding an asymptotic formula for the hyperfactorial is Problem 9.28 in Concrete Mathematics (2nd ed.). The answer they give uses Euler-Maclaurin, just as Aryabhata's answer does. They also mention that a derivation of the value of $C$ is in N. G. de Bruijn's Asymptotic Methods in Analysis, $\S$3.7.

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    @Moron: That sounds like it would work.2011-03-14
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Here is an asymptotic expression using EulerMcLaurin Summation.

\sum _{k=1}^{n} k \log k = \int_{1}^{n} x \log x\ \text{d}x + (n\log n)/2 + C' + (\log n + 1)/12+ \mathcal{O}(1/n^2)

$ = n^2(2 \log n - 1)/4 + (n\log n)/2 + (\log n)/12 + C + \mathcal{O}(1/n^2)$

for some constant $C$.

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    But you're right that the same procedure for finding Stirling's constant ought to work here. There's got to be a way to get this! :)2011-03-14