# 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 $${a}_{r} = \frac{1}{r}$$ 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 $$\frac{-log(1-x)}{1-x}$$?

Note by Kartik Sharma
3 years, 11 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 $$\gamma+\ln n$$, where $$\displaystyle \gamma = \int_{1}^{\infty} \frac{1}{\lfloor x \rfloor}-\frac{1}{x} dx$$ is the Euler-Mascheroni constant.

- 3 years, 11 months ago

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!

- 3 years, 11 months ago

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

Staff - 3 years, 11 months ago

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

- 3 years, 11 months ago

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

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

Staff - 3 years, 11 months ago