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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@user8250: It's submission or summation. – Mar 14 '11 at 16:12
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I may be asking a silly question, but what is a "submission"? – JavaMan Mar 14 '11 at 16:13
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@DJC srry :) it was a typo .... i meant sigma and i am reading more on how to use tags to write equations here.... – user8250 Mar 14 '11 at 16:14
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Just checking! :) – JavaMan Mar 14 '11 at 16:15
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3I doubt you will find a "closed form" formula. Maybe an asymptotic approximation? – Aryabhata Mar 14 '11 at 16:18
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5$log(1^1) + log(2^2) + log(3^3) + ... + log((n/2)^{n/2}) = log(1427*...)$? – The Chaz 2.0 Mar 14 '11 at 16:22
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Can I create an upper bound of it ?? I reached the same stage as @The Chaz – user8250 Mar 14 '11 at 16:25
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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.... – user8250 Mar 14 '11 at 16:44
2 Answers
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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.
Mike Spivey
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+1: Very nice ! :-) I am surprised the Plouffe inverter does not have this, though. – Aryabhata Mar 14 '11 at 20:53
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1@Moron: Thanks! It's satisfying to have finally figured out what that constant is. :) – Mike Spivey Mar 14 '11 at 20:55
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@Moron: It looks like Plouffe does have it (or a scaled version of it) after all. The numerical estimate I was using before wasn't precise enough. – Mike Spivey Mar 14 '11 at 20:59
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1I suppose one proof of this would be to start from $\zeta'(s) = -\sum \log k/k^s$ for $Re(s) \gt 1$ and extend it for all $s$ (which can be done using Euler McLaurin I believe), similar to the Riemann Zeta function. – Aryabhata Mar 14 '11 at 21:34
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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 $C$.
Aryabhata
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+1, although I get $C = \frac{1}{4}$, $(\log n)/12$ rather than $(\log n)/18$, and $O(\frac{1}{n^2})$. (And I have verified this numerically.) – Mike Spivey Mar 14 '11 at 17:03
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@MIke: You are correct. I have edited the answer. Thanks. That C=1/4 would need a proof, but that is a neat value for that constant. – Aryabhata Mar 14 '11 at 18:24
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@Moron: You're right to be suspicious of $C = 1/4$. It's not correct. The value of $C$ to six decimal places appears to be $0.248755$ - close to $1/4$ but not quite $1/4$. I'm not sure how to prove that or get an explicit expression, though. – Mike Spivey Mar 14 '11 at 19:23
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@Mike: Perhaps the Plouffe inverter has something. We would probably need more than 6 digits though. – Aryabhata Mar 14 '11 at 19:45
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I tried Plouffe, with the more precise estimate $0.24875448644$, but I didn't get anything exact. Just so you know, this number comes from using Mathematica to find the difference between the OP's sum and your estimate for large values of $n$. – Mike Spivey Mar 14 '11 at 19:58
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@Mike: There is an infinite series representation using the $B_{2k}$ which comes from the Euler McLaurin formula. It might be easier to use that to calculate the expression and might also lead us to a "closed" form expression. It is very similar (but not quite the same) to the series we get for $\sum \log k$ which gives us the Stirling's constant $\sqrt{2\pi}$, I believe. – Aryabhata Mar 14 '11 at 20:00
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I tried that, but I got $1/4 + \sum_{k=2}^{\infty} \frac{B_{2k}}{(2k)!} (2k-3)!$, which doesn't converge. I could have made a mistake, though. – Mike Spivey Mar 14 '11 at 20:03
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@Mike: You are right that it does not converge! I didn't actually try to work it out. Sorry about that. – Aryabhata Mar 14 '11 at 20:10
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1But 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! :) – Mike Spivey Mar 14 '11 at 20:14