Fwanads (pronounced fwah-nadz) are cute purple fuzzballs with one eye, two feet, and a striped tail. They are shy beasts who tend to exhibit a look of surprised dismay when rendered.
Fwanads are most active at night, multiply rapidly, and can be found in their native habitat at macmcrae.com.
These furry creatures share many characteristics in common with their mathematical counterparts, called FWANADs, an acronym that stands for Fractions with Adjacent Numerator and Denominator. Examples of FWANADS include
$$\frac{1}{2}, \frac{2}{3}, \frac{6}{5}, \frac{2014}{2013}$$Just like fwanads, these numbers have a positive disposition and love to multiply.
Products Of Fwanads
It’s not hard to see that any positive rational number can be obtained by multiplying together a finite number of FWANADs. For instance, we can write 3/7 as$$\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7}=\frac{3}{7}$$
Incidentally, this representation is not unique. Can you find another way to multiply FWANADs that also yields 3/7?
By multiplying FWANADs together in various ways, one can produce a remarkable variety of well-known mathematical quantities. For instance, consider the seemingly mundane process of multiplying a FWANAD by itself, such as in the product$$\frac{1001}{1000} \cdot \frac{1001}{1000} \cdot \frac{1001}{1000} \cdots \frac{1001}{1000}$$
where there are 1,000 terms in all. To six significant digits its value is 2.71692, which looks suspiciously close to e. It turns out that the larger the FWANAD one uses (where the number of factors is the value of the denominator), the closer the result comes to e≈2.71828. The symbol e is commonly thought to honor Leonhard Euler, in case you were wondering.
John Wallis, who was a contemporary of Euler, stumbled upon another unexpected FWANAD product in the process of using a creative technique for evaluating an integral to measure the area of a semicircle. He discovered that$$\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots=\frac{\pi}{2}$$
which represents our first instance of multiplying together infinitely many FWANADs.
Infinite products of FWANADs are delicate beasts. To make sense of such an expression, one multiplies together more and more of the individual terms, keeping track along the way of whether or not the overall product seems to approach a limiting value. To illustrate, consider the expression
$$\frac{1000}{1001} \cdot \frac{1001}{1002} \cdot \frac{1002}{1003} \cdot \frac{1003}{1004} \cdots$$Clearly the numbers being multiplied together are all close to 1, and getting closer the further we go. Therefore, one might expect the value of the infinite product to also be close to 1, or at least be positive. But upon further reflection, we realize that the overall value must tend toward 0, since at any stage the product reduces to a fraction of the form 1,000/N, where N becomes arbitrarily large. Keep this example in mind as we turn our attention to a final FWANAD product.
A Fwandical Square Root
There are many possible ways to write 2 as a product of FWANADs; for instance, any expression of the form$$\frac{N+1}{N} \cdot \frac{N+2}{N+1} \cdot \frac{N+3}{N+2} \cdots \frac{2 N}{2 N-1}$$
will do. There are $N$ terms in the product, so assuming that $N$ is even, we can remove every other term, say beginning with the first, in the hopes of obtaining $\sqrt2$ at least in the limit as $N$ goes to $∞$(By the way, Wallis is also credited with introducing the infinity symbol.) A quick computation gives$$\frac{1008}{1007} \cdot \frac{1010}{1009} \cdot \frac{1012}{1011} \cdots \frac{2012}{2011} \approx 1.41404$$
which looks quite promising.
To establish that$$\frac{N+2}{N+1} \cdot \frac{N+4}{N+3} \cdots \frac{2 N}{2 N-1} \rightarrow \sqrt{2}$$
in the limit as $N→∞$, we need only argue that the product of the removed terms is essentially the same as the product of the remaining terms, since taken together, the overall product is exactly 2. In other words, we must show that$$\frac{N+1}{N} \cdot \frac{N+3}{N+2} \cdots \frac{2 N-1}{2 N-2} \approx \frac{N+2}{N+1} \cdot \frac{N+4}{N+3} \cdots \frac{2 N}{2 N-1}$$Moving all terms to the left-hand side gives$$\frac{(N+1)^{2}}{N(N+2)} \cdot \frac{(N+3)^{2}}{(N+2)(N+4)} \cdots \frac{(2 N-1)^{2}}{(2 N-2)(2 N)} \approx 1$$
Every factor in the product is close to 1, but by now we know better than to accept this as conclusive evidence that the entire product will be approximately 1. Our numerical computation has us convinced that this time the product really does approach 1 as $N$ grows larger, but how can we prove it?
A Mathematical Triumph
The solution employs three elegant tools from analysis: a logarithm, an inequality, and a telescoping sum. Logarithms take products to sums, which are usually more manageable, so we begin by taking the natural log of the above expression. We now must show that the sum$$\ln \left(\frac{(N+1)^{2}}{N(N+2)}\right)+\cdots+\ln \left(\frac{(2 N-1)^{2}}{(2 N-2)(2 N)}\right)$$
is approximately 0. It helps to notice that the fractions in this sum are actually FWANADs, allowing us to write$$\ln \left(\frac{(N+1)^{2}}{N(N+2)}\right)=\ln \left(1+\frac{1}{N(N+2)}\right)$$
and similarly for the other terms.
The next step is to obtain a bound for this sort of expression. In other words, we know that so just how big could become? The answer is provided by a graph.
The tangent line to $y=\ln(1+x)$ at the origin is simply $y=x$. Since the graph of $\ln (1+x)$ is concave down, it must lie entirely below the tangent line, demonstrating that $\ln (1+x)< x$ for all positive real numbers $x$. Therefore,$$\ln \left(1+\frac{1}{N(N+2)}\right)<\frac{1}{N(N+2)}$$
and similarly for the other terms. Hence, the entire sum above is less than$$\frac{1}{N(N+2)}+\frac{1}{(N+2)(N+4)}+\cdots+\frac{1}{(2 N-2)(2 N)}$$
To simplify such a sum it is often advantageous to split up each term using partial fractions. In our case, we find that$$\frac{1}{x(x+2)}=\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+2}\right)$$
Therefore, the above sum becomes$$\frac{1}{2}\left(\frac{1}{N}-\frac{1}{N+2}+\frac{1}{N+2}-\frac{1}{N+4}+\cdots-\frac{1}{2 N}\right)$$
which neatly telescopes to just$$\frac{1}{2}\left(\frac{1}{N}-\frac{1}{2 N}\right)=\frac{1}{4 N}$$
This quantity clearly goes to 0 as N grows larger, which shows that the original product does indeed approach 1. And this means that$$\frac{N+2}{N+1} \cdot \frac{N+4}{N+3} \cdots \frac{2 N}{2 N-1} \rightarrow \sqrt{2}$$
In the limit as and we’re done.
By Sam Vandervelde