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
5 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.

- 5 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!

- 5 years, 11 months ago

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

Staff - 5 years, 11 months ago

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

- 5 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 - 5 years, 11 months ago