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 November 2nd, 2019, 12:16 PM #1 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle Generalization of Harmonic Series $h(x)$ is a generalized $H_n$, and $H_n = \sum\limits_{k=1}^n \dfrac{1}{k}$. Assumptions:For $n \in \mathbb{N}$, $h(n)=H_n$; $h(1)=1$; $h(x)=h(x-1) + \dfrac{1}{x}$. I started by $$\Large{f(n)= \int_1^n h(z)dz}$$ and at next page I derived at $$\Large{\int_1^n h(z) dz = \ln (n!) + (n-1)k},$$ where $$\large{k=\int_0^1 h(z)dz},$$ which is going to be called Euler-Mascheroni constant. Now, my brain stopped working, so what should be done after $$\Large{\int_1^n h(z) dz = \ln (n!) + (n-1)k}.$$ Thanks from idontknow November 2nd, 2019, 12:32 PM #2 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle I can just assume $\int h(z) dz = \ln (z!) + (z-1)k$ and check if it holds true for $\int_1^n h(z) dz$.     November 2nd, 2019, 07:41 PM #3 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle How can I show $$\large{\int_0^1 h(z)dz}$$ is convergent and between 0 and 1, or it is equal to $$\large{\lim\limits_{n \rightarrow \infty} H_n - \ln (n)}$$? November 3rd, 2019, 02:12 PM   #4
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Quote:
 Originally Posted by tahirimanov19 How can I show $$\large{\int_0^1 h(z)dz}$$ is convergent and between 0 and 1, or it is equal to $$\large{\lim\limits_{n \rightarrow \infty} H_n - \ln (n)}$$?
Knowing that $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-z}},$$

I found,
$$\Large{\int_{0}^1 h(z) dz = \int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx}.$$

What to do next??? November 3rd, 2019, 03:37 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,107 Thanks: 2324 Knowing that what?? November 3rd, 2019, 05:46 PM #6 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle Harmonic series... November 3rd, 2019, 05:48 PM   #7
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Math Focus: Area of Circle
Quote:
 Originally Posted by tahirimanov19 Knowing that $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-x}} dx,$$ I found, $$\Large{\int_{0}^1 h(z) dz = \int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx}.$$ What to do next???
I made a mistake... It is $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-x}} dx,$$ November 4th, 2019, 02:31 AM #8 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math $\displaystyle H_n =\ln(n) +\gamma +\epsilon_n .$ All of integrals are correct , the last one maybe has to do with the unknown Last edited by idontknow; November 4th, 2019 at 02:33 AM. Tags generalization, harmonic, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Elementary Math 2 September 17th, 2019 01:38 AM William Labbett Number Theory 0 October 11th, 2014 04:57 AM Daltohn Calculus 3 March 2nd, 2014 01:31 PM julian21 Real Analysis 2 December 8th, 2010 01:38 PM brunojo Real Analysis 11 December 2nd, 2007 08:49 AM

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