Generalization of Harmonic series and Euler-Mascheroni Constant

Question

It is like a two-part question, and if answers already exist on Stack, would you be kind enough to add links to it.

First is a generalization of the Harmonic series. I made three assumptions:

For $n \in \mathbb{N}$, $h(n)=H_n \\$;
$h(1)=1 \\$;
$h(z)=h(z-1)+ \dfrac{1}{z}$.

And I let $f(n)$ be $\int\limits_1^n h(z) dz$. At the end, I found $$\int\limits_1^n h(z) dz = \ln (n!) + (n-1) \cdot \int\limits_{0}^1 h(z) dz.$$

My problem is with $\int\limits_{0}^1 h(z) dz$. Is there a way to show that it is equal to $\lim\limits_{n \rightarrow \infty} H_n - \ln (n)$? Or how can I show that $\int\limits_{0}^1 h(z) dz$ is convergent and equal to some number between 0 and 1?

Thank you, in advance.

Ali Shadhar · Accepted Answer · 2019-11-03T22:14:18.463

1

Starting with the definition of the harmonic number

$$H_z=\int_0^1\frac{1-x^z}{1-x} \ dx$$

we can write

$$\int_0^1 H_z \ dz=\int_0^1\frac1{1-x}\left(\int_0^1(1-x^z)\ dz\right)\ dx\\=\int_0^1\frac1{1-x}\left(\frac{1-x+\ln x}{\ln x}\right)\ dx\\=\int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx\\=\gamma=\lim_{n\to \infty}\left(H_n-\ln n\right)$$

where $\gamma$ is the Euler constant

About the last limit, you can find variant proofs here.

edited Nov 03 '19 at 22:14

answered Nov 03 '19 at 05:20

Ali Shadhar

25,498

How do you jump from the 3rd line to the 4th? – Tahir Imanov Nov 03 '19 at 18:40
1

@TahirImanov that's the definition of euler constant in integral and limit representation. It was proved here on MSE I'll mention the link when I find it. – Ali Shadhar Nov 03 '19 at 18:43
Ok. Thanks for the answer. – Tahir Imanov Nov 03 '19 at 18:47
@TahirImanov you welcome and thanks for the upvote :) – Ali Shadhar Nov 03 '19 at 18:54
Here is the link: https://math.stackexchange.com/questions/344314/showing-that-lim-n-to-infty-sumn-k-1-frac1k-lnn-0-5772-ldots/344574#344574 – Tahir Imanov Nov 03 '19 at 19:54
@TahirImanov yes that's the one.. allow me to use the link. – Ali Shadhar Nov 03 '19 at 20:56

Ali Shadhar · Answer 2 · 2019-11-04T00:08:38.270

Or we can start with the series representation of the harmonic number:

$$H_z=\sum_{k=1}^z\frac1k=\lim_{n\to \infty}\sum_{k=1}^n\left(\frac1k-\frac1{z+k}\right)$$

Then

$$\int_0^1 H_z dz=\lim_{n\to \infty}\sum_{k=1}^n\int_0^1\left(\frac1k-\frac1{z+k}\right)\ dz\\=\lim_{n\to \infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=1}^n(\ln(1+k)-\ln(k))\right)\\=\lim_{n\to \infty}\left(H_n-\ln(1+n)\right), \quad 1+n\to n\\=\lim_{n\to \infty}\left(H_{n-1}-\ln(n)\right)\\=\lim_{n\to \infty}\left(H_{n}-\frac1n-\ln(n)\right)\\=\lim_{n\to \infty}\left(H_{n}-\ln(n)\right)$$

Note that $\sum_{k=1}^n(\ln(1+k)-\ln(k))=\ln(1+n)$ follows from using the telescoping series.

Generalization of Harmonic series and Euler-Mascheroni Constant

2 Answers2