I was wondering on a problem in Alan Tucker's Combinatorics when I noticed this and I wanted to know if it is true(Though I have checked on Wolfram Alpha and it is quite true).

The generating function for \({a}_{r} = \frac{1}{r}\) is of course \(-log(1-x)\).

Therefore, \(h(x) = -log(1-x)\)

Hence, \(h*(x) = \frac{-log(1-x)}{1-x}\) , where \( h * (x) \) is the generating function for the sums of ith coefficient of \( h(x) \).

As a result, the co-efficient of \({x}^{n}\) in \(h*(x)\) is equal to \(1 + \frac{1}{2} + \frac{1}{3}+.... + \frac{1}{n}\).

If this is true, is there any way to find the coefficient of \({x}^{n}\) in \(\frac{-log(1-x)}{1-x}\)?

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## Comments

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TopNewestYou mean a closed form? I'm afraid I don't know of one (it probably does not exist). However, the asymptotic limit is \(\gamma+\ln n\), where \(\displaystyle \gamma = \int_{1}^{\infty} \frac{1}{\lfloor x \rfloor}-\frac{1}{x} dx\) is the Euler-Mascheroni constant.

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Are you sure this is true? *I haven't heard of anything like that. BTW, try finding the coefficient of the \(h(x)\) too(with proof, if you can). Thanks for that info!

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@Calvin Lin @brian charlesworth @Jon Haussmann

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By \( h * (x) \), do you mean \( h' (x) \)?

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No sir, by h*(x), I meant the generating function of the sums of the \({a}_{r}\)s.

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Thanks. I added it in to clarify what that terminology is.

Usually for such Maclaurin expansions, you simply multiply the different parts together.

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