My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree2Thanks
  • 1 Post By tahirimanov19
  • 1 Post By tahirimanov19
Reply
 
LinkBack Thread Tools Display Modes
November 2nd, 2019, 12:16 PM   #1
Senior Member
 
tahirimanov19's Avatar
 
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
tahirimanov19 is offline  
 
November 2nd, 2019, 12:32 PM   #2
Senior Member
 
tahirimanov19's Avatar
 
Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Red face

I can just assume $\int h(z) dz = \ln (z!) + (z-1)k$ and check if it holds true for $\int_1^n h(z) dz$.
tahirimanov19 is offline  
November 2nd, 2019, 07:41 PM   #3
Senior Member
 
tahirimanov19's Avatar
 
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 $$\large{\lim\limits_{n \rightarrow \infty} H_n - \ln (n)}$$?
tahirimanov19 is offline  
November 3rd, 2019, 02:12 PM   #4
Senior Member
 
tahirimanov19's Avatar
 
Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Quote:
Originally Posted by tahirimanov19 View Post
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???
tahirimanov19 is offline  
November 3rd, 2019, 03:37 PM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 21,107
Thanks: 2324

Knowing that what??
skipjack is online now  
November 3rd, 2019, 05:46 PM   #6
Senior Member
 
tahirimanov19's Avatar
 
Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Harmonic series...
tahirimanov19 is offline  
November 3rd, 2019, 05:48 PM   #7
Senior Member
 
tahirimanov19's Avatar
 
Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
Quote:
Originally Posted by tahirimanov19 View Post
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 $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-x}} dx,$$
Thanks from idontknow
tahirimanov19 is offline  
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.
idontknow is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
generalization, harmonic, series



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Harmonic series idontknow Elementary Math 2 September 17th, 2019 01:38 AM
harmonic series William Labbett Number Theory 0 October 11th, 2014 04:57 AM
Harmonic series Daltohn Calculus 3 March 2nd, 2014 01:31 PM
Harmonic Series julian21 Real Analysis 2 December 8th, 2010 01:38 PM
Harmonic series brunojo Real Analysis 11 December 2nd, 2007 08:49 AM





Copyright © 2019 My Math Forum. All rights reserved.