# A lovely observation

Posted by: Gary Ernest Davis on: January 13, 2012

Ben Vitale (@BenVitale on Twitter) recently made the elementary and lovely observation that

$\frac{1+3}{5+7}=\frac{1+3+5}{7+9+11}=\frac{1+3+5+7}{9+11+13+15}= \ldots = \frac{1}{3}$

x

Some people asked why this is so, and the answer, simple as it is, makes a nice middle or high school problem

First let’s think about what’s happening here.

The numerators of these fractions are sums of the first few odd numbers.

The denominators are sums of the same number of odd numbers, starting where the numerator leaves off.

To use variable notation – something some middle schoolers are still struggling with – the numerators look like:

$1+3+5=\ldots + 2k-1$

as the positive integer $k$ increases by 1.

The denominators, notationally a little more complicated, look like:

$(2k+1)+(2k+3)+\ldots + (2k+2k-1)$

We can find short algebraic expressions for these numerators and denominators, and it is these simple expressions that will show us where the fraction $\frac{1}{3}$ comes from.

Let’s write $S(k)=1+3+5+\ldots +2k-1$

and

$T(k)=2+4+6+\ldots +2k-2$

x

Then

$S(k)+T(k)= 1+2+3+\ldots + 2k-2 + 2k-1= \frac{1}{2}(2k-1)\times 2k=k(2k-1)=2k^2-k$

and

$T(k)=2+4+6+\ldots +2k-2= 2(1+2+3+\ldots + (k-1))=2\times \frac{1}{2}(k-1)k=k^2-k$

Therefore, $S(k)=(2k^2-k)-(k^2-k)=k^2$.

For the denominator we have:

$(2k+1) + (2k+3) + \ldots + (2k + 2k-1) = (2k+2k+\ldots +2k) + (1+3+\ldots +2k-1)= 2k^2+k^2=3k^2$

Therefore,

$\frac{1+3+5+\ldots +2k-1}{(2k+1)+(2k+3)+\ldots + (2k+2k-1)}=\frac{k^2}{3k^2}=\frac{1}{3}$

independent of $k$.

The situation for similar sums of even numbers is not quite so simple.

$2+4+6+\ldots + 2k = 2(1+2+3+\ldots + k)=2\times \frac{1}{2}k(k+1)=k^2+k$

while

$(2k+2)+(2k+4)+(2k+6)+\ldots +(2k+2k) = (2k+2k+2k+\ldots +2k)+(2+4+6+\ldots +2k)= 2k^2+k^2+k=3k^2+k$

x

Therefore,

$\frac{2+4+6+\ldots +2k}{(2k+2)+(2k+4)+\ldots + (2k+2k)}=\frac{k^2+k}{3k^2+k}=\frac{1+1/k}{3+1/k}\to \frac{1}{3}$

x

as $k$ increases.

A lovely observation, some simple algebra, and a challenging yet rewarding problem for middle and high school students.

### 3 Responses to "A lovely observation"

I see it as being a consequence of the squares being the sum of the odd numbers, and I might approach that with an idea more like http://en.wikipedia.org/wiki/File:Proofwithoutwords.svg .

We have (sum of the first k odd numbers) / (sum of the next k odd numbers).

Sum of the first k odd numbers is k^2.

And sum of the next k odd numbers = (sum of the first 2k odd numbers) – (sum of the first k odd numbers) = (2k)^2 – k^2 = 3k^2.

— Joshua thanks for this comment. Apart from the visual argument, isn’t this exactly what I said? Gary Davis

Mathematically these ideas are identical, but visually and comprehensively Josh’s method is much easier to follow, tho is uses the property of the sum of odds as a shortcut to the answer. But personally that’s where my mind went.

An easy extension of this beautiful observation is that the increment need not be two.

In symbols:

(1 + 1 + k + 1 + 2k + . . . + 1 + nk)/(3 + nk + 3 + nk + k + 3 + nk + 2k + . . . + 3 + nk + nk) = 1/3

for any value of k for which the denominator is not zero.

In other words, the sum of the first n evenly spaced quantities, starting with 1, divided by the sum of the next n evenly spaced quantities, starting from two more than the last term of the numerator, is equal to 1/3. The “evenly spaced” amount need not be an integer.