Difference between revisions of "Harmonic series"

(Proof of harmonic series converging)
m (How to solve)
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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>\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=\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>
 
 
We could have also shown that the harmonic series diverges by showing that <math>\displaystyle\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=1</math> and that the limit must be less than 1 for it to converge.
 

Revision as of 10:20, 23 August 2006

There are several types of harmonic series.

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

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

The general harmonic series, $\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}$, has its value depending on the value of the constants $a$ and $b$.

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 converges 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 $\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=(1+\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty$

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