Difference between revisions of "Harmonic series"
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The alternating harmonic series, | 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>\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>. | ||
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The [[zeta-function]] is a harmonic series when the input is one. | The [[zeta-function]] is a harmonic series when the input is one. | ||
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== How to solve == | == How to solve == | ||
− | ===Harmonic Series=== | + | === Harmonic Series === |
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 | 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}{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> | <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=== | + | === Alternating Harmonic Series === |
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+ | === General Harmonic Series === | ||
+ | <math>\displaystyle\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> | ||
+ | |||
+ | <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> | ||
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+ | '''Case 2:''' <math>a<b</math> | ||
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+ | <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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Thus, <math>\sum_{i=1}^{\infty}\frac{1}{ai+b}=\infty</math> |
Revision as of 13:13, 12 November 2006
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 This sum slowly approaches infinity.
The alternating harmonic series, , though, approaches .
The zeta-function is a harmonic series when the input is one.
Contents
[hide]How to solve
Harmonic Series
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, is greater than . We also know that the third and and fourth terms, and , add up to something greater than . And we continue grouping the terms between powers of two. So we have
Alternating Harmonic Series
General Harmonic Series
is the general harmonic series, where each term is the reciprocal of a term in an arithmetic series.
Case 1:
Case 2:
Thus,