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Compute

$$\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $$ where $\gamma$ - Euler's constant.

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as always: what is the source of the problem, and what sort of material is in the book just before it? –  Will Jagy Jan 15 '13 at 17:03
    
@WillJagy: it comes from a collection of calculus problems of all kinds that I and my colleagues have gathered. I don't know what is the source of it. –  Chris's sis Jan 15 '13 at 17:13
    
Please, $H_k=?$ –  Elias Jan 15 '13 at 17:23
    
@Elias: mathworld.wolfram.com/HarmonicNumber.html –  Chris's sis Jan 15 '13 at 17:24
    
Mathematica calculated it to $ \pi^2 \over 12 $ –  Santosh Linkha Jan 15 '13 at 17:31
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1 Answer

up vote 8 down vote accepted

We have \begin{align} 2\sum_{k=1}^n \frac{H_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}, \text{ swapping the order of summation on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=k}^n \frac{1}{jk}, \text{ changing variables on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^n \frac{1}{jk} + \sum_{k=1}^n \frac{1}{k^2} \\ &= \left(\sum_{k=1}^n \frac{1}{k} \right)^2 + \sum_{k=1}^n \frac{1}{k^2} \\ &= H_n^2+ H^{(2)}_n. \\ \end{align}

Thus \begin{align*} \lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\log n+\gamma)^2\right) &= \lim_{n\to\infty} \frac{1}{2}\left(H_n^2+ H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \lim_{n\to\infty} \frac{1}{2}\left((\log n + \gamma)^2 + O(\log n/n) + H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \frac{1}{2}\lim_{n\to\infty} \left( H^{(2)}_n + O(\log n/n) \right) \\ &= \frac{1}{2}\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2}\\ &= \frac{\pi^2}{12}. \end{align*}

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Beautiful manipulation. Thanks (+1) –  Chris's sis Jan 15 '13 at 17:36
    
@Chris'ssister: The manipulation in the first part (getting to $H^2_n + H^{(2)}_n$) is a trick that works any time you're essentially adding up the upper (or lower) triangular part of a symmetric square array of numbers. The details are in Concrete Mathematics, pp. 36-37. It's a useful trick to know. –  Mike Spivey Jan 15 '13 at 17:38
    
Nice! Thanks again for the details. –  Chris's sis Jan 15 '13 at 17:41
    
did you think to publish a book with multiseries problems and solutions? I also noticed in one of my previous posts some very nice manipulations with double sums you did. –  Chris's sis Jan 15 '13 at 18:14
    
@Chris'ssister: That would be an interesting book, but, no, I don't have any plans to write such a one right now. A lot of what I know about manipulating sums I learned from practice and Concrete Mathematics. I highly recommend that book. Its exercises make great practice problems, too. –  Mike Spivey Jan 15 '13 at 18:33
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