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In a recent blog A Lovely Observation Gary Davis (@RepublicOfMath) elaborated on an observation of Ben Vitale (@BenVitale) to the effect that

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

In the fractions both numerators and denominators are sums of successive odd numbers: the numerators start with 1, the denominators where the numerators leave off. Thus naturally derivation of the formula for the sum of successive odd numbers is the key to the explanation of Ben Vitale's observation. There are ways and ways to obtain the required expression. The most elementary one is to observe that

$$S(n) = 1 + 3 + \ldots + (2n - 1)$$

is the sum of an arithmetic sequence ,
where and . With a nod to the young F. Gauss,

$$a+ (a + d)+(a + 2d)+ \ldots + [a + (n-1)d] = \frac{n(2a + (n-1)d)}{2},$$

which in the case of odd numbers gives

$$1+3+ \ldots + (2n-1) = \frac{n(2 + (n-1)2)n}{2}=n^2.$$

Another way to derive that formula is to noticee that the sum of odd numbers is the sum of all numbers less the sum of the even ones. It is convenient at this point to start using the symbol of to make the formulas shorter and more manageable:

$$S(n)=\sum_{k=1}^{n}(2k-1)=n^2$$

and, for the sum of all integers from to ,

$$N(n)=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$$

Now a little of summation magic:

$$\begin{align}S(n) &= N(2n) - 2N(n) = \sum_{k=1}^{2n}k - 2\sum_{k=1}^{n}k \\ &= \sum_{k=1}^{2n}k - \sum_{k=1}^{n}(2k)\\& = \frac{2n(2n+1)}{2}-2\frac{n(n+1)}{2}=(2n^{2}+n)-(n^{2}+n)=n^2,\end{align}$$

with the same result. Let's note that here we made a subtle use of the homogeneity property of function , namely . Simple as it appears, it was central to the derivation. For Ben Vitale's observation we'll have to do something very similar:

$$\begin{align}(2n+1)+(2n+3)+\ldots +(2n + (2n-1)) &= \sum_{k=1}^{2n}(2k-1)-\sum_{k=1}^{n}(2k-1) \\&= (2n)^2-n^2=3n^2.\end{align}$$

Ben Vitale's observation comes to

$$\frac{1+2+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(4n-1)}=\frac{n^2}{3n^2}=\frac{1}{3}.$$

A generalization of this would be to take, say, twice as many terms in the denominator as in the numerator. This can be done in more than one way:

$$\frac{1+3+\ldots +(2n-1)}{1+3+\ldots +(4n-1)}=\frac{1}{4}$$

or

$$\frac{1+3+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(6n-1)}=\frac{1}{8},$$

because , as before. Going further, there are at least three ways to have three times as many terms in the denominator as in the numerator:

$$\frac{\sum_{k=1}^{n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)},\space \frac{\sum_{k=n+1}^{2n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)},\space \frac{\sum_{k=2n+1}^{3n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)}.$$

One can easily check that the three fractions are equal to , , and , respectively.

Obviously, our success in getting a simple expression is due to the homogeneity of function : . (There is a difference between the two functions and . For the obvious reasons, the former is said to be homogeneous of order (degree) 1, the later of order 2.)

As Gary Davis notes at the end of his post, this kind of algebraic manipulations is accessible to middle and high school students. An additional example of the infinite series of the reciprocals of squares requires no more effort but a little hand-waving if not presented at the beginning Calculus class.

Assume we know that

$$\sum_{k=1}^{\infty}k^{-2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots =\frac{\pi^2}{6}.$$

Then of course, for the series that involves only even numbers,

$$\sum_{k=1}^{\infty}(2k)^{-2}=\frac{1}{2^2}+\frac{1}{(2^{2})(2^{2})}+\frac{1}{(2^{2})(3^{2})}+\ldots =\frac{\pi^2}{4\cdot 6}=\frac{\pi^2}{24}.$$

As a reward for the effort, we get an expression for the series of the reciprocals of the squares of odd numbers:

$$\sum_{k=1}^{\infty}(2k-1)^{-2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$

Who would have thought that the series of the reciprocals of odd squares sums up to three times a similar series for even squares!