Difference between revisions of "Harmonic series"

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

@@ Line 4: / Line 4: @@
 The the most basic harmonic series is the infinite sum
-<math>\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots</math>
+<math>\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots</math>
 This sum slowly approaches infinity.
 The alternating harmonic series,
-<math>\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though, approaches <math> \displaystyle \ln 2</math>.
+<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>.
-The general harmonic series, <math>\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}</math>, has its value depending on the value of the constants <math>a</math> and <math>b</math>.
 The [[zeta-function]] is a harmonic series when the input is one.
@@ Line 16: / Line 14: @@
 == How to solve ==
-===Harmonic Series===
+=== Harmonic Series ===
+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
+<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>
+=== Alternating Harmonic Series ===
+=== General Harmonic Series ===
+<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.
+'''Case 1:''' <math>a\ge b</math>
+<math>ai+a\ge ai+b</math>
+<math>\frac{1}{ai+b}\ge\frac{1}{ai+a}=\frac{1}{a}\left(\frac{1}{i+1}\right)</math>
+<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>
+'''Case 2:''' <math>a<b</math>
+<math>ai+b<bi+b</math>
+<math>\frac{1}{ai+b}>\frac{1}{bi+b}=\frac{1}{b}\left(\frac{1}{i+1}\right)</math>
-It can be shown that the harmonic series converges by grouping the terms.  We know that the first term, 1, 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
+<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>
-<math>\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=(1+\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty</math>
-===Alternating Harmonic Series===
+Thus, <math>\sum_{i=1}^{\infty}\frac{1}{ai+b}=\infty</math>

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