# Infinite Series That Equal Two: So a mathematician walks into a bar…

Infinity is certainly a strange mathematical concept.  It is not like many of the numbers that people are used to dealing with on a day to day basis.  However, it is extremely helpful to have as a tool for the average physicist.  Today I want to pass a math joke off to you that will help you understand some of the most basic of infinite series.

First off, what is an infinite series?

Let’s say I add 1 to 1 one time.  This is represented as 1+1 and is equal to 2.  If I add 1 to 1 two times then I can say 1+1+1=3.  If I add 1 to 1 an infinite amount of times I get 1+1+1+…= ? because this series does not converge!  There are several tests in calculus and other kinds of mathematics to help thinkers and scientists of all types discern if different kinds of series and additions converge.  Rather than get bogged down in the convergence of this moment and talking about how these things can be proven let us take a look instead at an example of a series that does converge.

If I add $\frac{1}{2}$ + $\frac{1}{4}$ I will get $\frac{3}{4}$.  This can be shown as $\frac{1}{2}$ + $\frac{1}{4}$ = $\frac{2}{4}$ + $\frac{1}{4}$ $\frac{2+1}{4}$ = $\frac{3}{4}$.  If I add $\frac{1}{8}$ to that, or half of what I added the last time [ $\frac{1}{2}$* $\frac{1}{4}$ = $\frac{1}{4*2}$ = $\frac{1}{8}$], we will see that $\frac{3}{4}$ + $\frac{1}{8}$ $\frac{6}{8}$ + $\frac{1}{8}$ = $\frac{1+6}{8}$ = $\frac{7}{8}$.  So exciting!  Now let’s do this again, and this time see if you notice the pattern. $\frac{7}{8}$ + $\frac{1}{16}$ = $\frac{14}{16}$ $\frac{1}{16}$ = $\frac{14+1}{16}$ = $\frac{15}{16}$.  If you haven’t spotted the pattern yet and want to do so on your own, try to add the next value ( $\frac{1}{32}$) and see if you spot it!  Spoilers are in the next paragraph.

As we continue to do this an infinite amount of times each addition increases the denominator, the number in the bottom of the fraction, by a power of 2 ( $2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4=16$ $2^n$).  However the numerator, or the upper number of the fraction, is always the denominator minus 1.  This means as we do this for the n = 1000000 th = $10^6$ time the numerator will be $(2^n)-1$ while the denominator is simply $2^n$. Do this an infinite amount of times and $\lim(n \to \inf)\frac{2^n-1}{2^n} \approx \frac{2^n}{2^n} = 1$.  This is as close to the value of 1 as one can truly get.

Now for a math joke that you can share yourself!

A mathematician walks into a bar.  They order a beer.  They are followed by a second mathematician who orders $\frac{1}{2}$ of a beer who is followed by another mathematician who orders half of the last mathematicians order ( $\frac{1}{4}$ of a beer) who is followed by an infinite amount of mathematicians who order half of the last order before them.  The bartender gets flustered and pours two drinks. $1 + \Sigma_{n=1}^{n=\inf} (\frac{1}{2})^n = \Sigma_{n=0}^{n=\inf} (\frac{1}{2})^n = 2$

If you are looking for me to Taylor more series for your calculus needs I would be infinitely delighted to continue this line of thinking!  Look for future posts regarding reflections, absorptions, and polarization in the upcoming weeks for how this concept can be related to different aspects of physics!

Best,

Josh