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Difference between revisions of "Harmonic series"

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A '''harmonic series''' is a form of the zeta function :
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Generally, a '''harmonic series''' is a [[series]] whose terms involve the [[reciprocal]]s of the [[positive integer]]s.
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<math> \zeta (x) = 1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+... </math>.
 
  
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When <math>\ x</math> has a value less than or equal to one the function outputs infinity.  Euler found that when <math>\ x=2</math>, the zeta function outputs <math>\frac{\pi^2}{6} </math>.
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There are several sub-types of '''harmonic series'''.
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Euler also realized that since every number is the multiplication of some order of primes, then the zeta function is equal to <math>(1+\frac{1}{2^x}+\frac{1}{4^x}+...)(1+\frac{1}{3^x}+\frac{1}{9^x}+...)...(1+\frac{1}{p^x}+\frac{1}{(p^2)^x}+...)...</math>
 
  
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Riemann found that when complex numbers are the input to the zeta function, the resulting graph is that which aids in the finding of the exact value of <math>\ \pi (n)</math> or the number of primes less than or equal to <math>\ n</math>.
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The the most basic harmonic series is the infinite sum
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<math>\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots</math>
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This sum slowly approaches infinity.
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The alternating harmonic series,
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<math>\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though, approaches <math>\ln 2</math>.
  
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The [[zeta-function]] is a harmonic series when the input is one.
  
 
== How to solve ==
 
== How to solve ==
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=== Harmonic Series ===
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It can be shown that the harmonic series diverges by grouping the terms.  We know that the first term, <math>1</math>, added to the second term, <math>\frac{1}{2}</math> is greater than <math>\frac{1}{2}</math>.  We also know that the third and and fourth terms, <math>\frac{1}{3}</math> and <math>\frac{1}{4}</math>, add up to something greater than <math>\frac{1}{2}</math>.  And we continue grouping the terms between powers of two.  So we have
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<math>\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty</math>
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=== Alternating Harmonic Series ===
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=== General Harmonic Series ===
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<math>\sum_{i=1}^{\infty}\frac{1}{ai +b}</math> is the general harmonic series, where each term is the reciprocal of a term in an arithmetic series.
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'''Case 1:''' <math>a\ge b</math>
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<math>ai+a\ge ai+b</math>
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<math>\frac{1}{ai+b}\ge\frac{1}{ai+a}=\frac{1}{a}\left(\frac{1}{i+1}\right)</math>
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<math>\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{a} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty</math>
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'''Case 2:''' <math>a<b</math>
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<math>ai+b<bi+b</math>
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<math>\frac{1}{ai+b}>\frac{1}{bi+b}=\frac{1}{b}\left(\frac{1}{i+1}\right)</math>
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<math>\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{b} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty</math>
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Thus, <math>\sum_{i=1}^{\infty}\frac{1}{ai+b}=\infty</math>

Latest revision as of 19:52, 13 March 2022

Generally, a harmonic series is a series whose terms involve the reciprocals of the positive integers.

There are several sub-types of harmonic series.

The the most basic harmonic series is the infinite sum $\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ This sum slowly approaches infinity.

The alternating harmonic series, $\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ , though, approaches $\ln 2$.

The zeta-function is a harmonic series when the input is one.

How to solve

Harmonic Series

It can be shown that the harmonic series diverges by grouping the terms. We know that the first term, $1$, added to the second term, $\frac{1}{2}$ is greater than $\frac{1}{2}$. We also know that the third and and fourth terms, $\frac{1}{3}$ and $\frac{1}{4}$, add up to something greater than $\frac{1}{2}$. And we continue grouping the terms between powers of two. So we have $\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty$

Alternating Harmonic Series

General Harmonic Series

$\sum_{i=1}^{\infty}\frac{1}{ai +b}$ is the general harmonic series, where each term is the reciprocal of a term in an arithmetic series.

Case 1: $a\ge b$

$ai+a\ge ai+b$

$\frac{1}{ai+b}\ge\frac{1}{ai+a}=\frac{1}{a}\left(\frac{1}{i+1}\right)$

$\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{a} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty$

Case 2: $a<b$

$ai+b<bi+b$

$\frac{1}{ai+b}>\frac{1}{bi+b}=\frac{1}{b}\left(\frac{1}{i+1}\right)$

$\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{b} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty$

Thus, $\sum_{i=1}^{\infty}\frac{1}{ai+b}=\infty$

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