Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Harmonic series"

(Proof of harmonic series converging)
m (→‎How to solve)
Line 17: Line 17:
  
 
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 09:20, 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 $\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=(1+\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Harmonic_series&oldid=9821"