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 $\frac{-log(1-x)}{1-x}$ ?

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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.

Jake Lai - 6 years, 4 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!

Kartik Sharma - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

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

Kartik Sharma - 6 years, 4 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.

Calvin Lin Staff - 6 years, 4 months ago

@Calvin Lin @brian charlesworth @Jon Haussmann

Kartik Sharma - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$