Harmonic series sum

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 ar=1r{a}_{r} = \frac{1}{r} is of course log(1x)-log(1-x).

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

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

As a result, the co-efficient of xn{x}^{n} in h(x)h*(x) is equal to 1+12+13+....+1n1 + \frac{1}{2} + \frac{1}{3}+.... + \frac{1}{n}.

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

Note by Kartik Sharma
4 years, 5 months ago

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You mean a closed form? I'm afraid I don't know of one (it probably does not exist). However, the asymptotic limit is γ+lnn\gamma+\ln n, where γ=11x1xdx\displaystyle \gamma = \int_{1}^{\infty} \frac{1}{\lfloor x \rfloor}-\frac{1}{x} dx is the Euler-Mascheroni constant.

Jake Lai - 4 years, 5 months ago

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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)h(x) too(with proof, if you can). Thanks for that info!

Kartik Sharma - 4 years, 5 months ago

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

Calvin Lin Staff - 4 years, 5 months ago

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

Kartik Sharma - 4 years, 5 months ago

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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.

Calvin Lin Staff - 4 years, 5 months ago

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