Astounding “Natural” Identities

The modest sequence $$1,2,3,4,\ldots$$ can do some rather awe-inspiring things, when properly arranged. Here’s a short list of some of its many impressive feats.

There are numerous expressions for $$\pi$$ relying on the progression of integers, including the Wallis formula: $$\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots$$ (which can be derived from Complex Analysis using an infinite product representation for the sine function) and an elegant alternating sum: $$\frac{\pi-3}{4} = \frac{1}{2\cdot 3\cdot 4}-\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}-\cdots$$ (try to prove this!). Euler’s number $$e$$ has similarly surprising formulas, such as $$\frac{1}{e-2} = 1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\cdots}}}}$$ (listed at MathWorld) and $$\frac{e}{e-1} = 1+\frac{1+\frac{1+\frac{1+\cdots}{2+\cdots}}{2+\frac{2+\cdots}{3+\cdots}}}{2+\frac{2+\frac{2+\cdots}{3+\cdots}}{3+\frac{3+\cdots}{4+\cdots}}}$$ (which is problem 1745 in Mathematics Magazine, posed by Gerald A. Edgar).

The list doesn’t stop here! The nested square-root identity $$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$ is attributed to Ramanujan (on this Wikipedia page). As another curiosity, the sequence $$\frac{1}{2},\qquad \frac{1}{2} \Big/ \frac{3}{4},\qquad \frac{\frac{1}{2} \big/ \frac{3}{4}}{\frac{5}{6} \big/ \frac{7}{8}},\qquad \frac{\frac{1}{2} \big/ \frac{3}{4}}{\frac{5}{6} \big/ \frac{7}{8}} \bigg/ \frac{\frac{9}{10} \big/ \frac{11}{12}}{\frac{13}{14} \big/ \frac{15}{16}},\qquad\ldots$$ (which relates to the Thue-Morse sequence) can be shown to converge to $$\sqrt{2}/2$$.

There are many pretty/unexpected/crazy formulas obtainable from the natural numbers $$1,2,3,4,\ldots$$ that could not fit in this post. What are some of your favorites?

Some Strange Sums

If someone told you to compute the value of the infinite sum $$Z = 1+2+3+4+\cdots$$, you would tell them that it simply does not converge. But what if they insisted that you assign a finite value to this sum? Which value would you give? Is there any way to make sense of this question? Yes, and the answer is $$-1/12$$.

Before tackling this sum, let’s try a slightly easier one, known as Grandi’s Series: $$G = 1-1+1-1+\cdots$$. How can we compute its value? Let’s write \begin{align*}G &= 1-1+1-1+1-\cdots \text{ and}\\ G &= 0+1-1+1-1+\ldots,\end{align*} and adding these term by term gives $$2G = 1+0+0+0+0+\cdots = 1$$, i.e., $$G = 1/2$$. So we got a value! The same “trick” lets us sum $$H = 1-2+3-4+\cdots$$ like this: \begin{align*}2H &= 1 + (-2+1) + (3-2) + (-4+3) + \cdots\\ &= 1-1+1-1+\cdots\\ &= G = 1/2,\end{align*} so $$H = 1/4$$.

What’s really going on here? A good way to understand it is to look at the functions $$g(x) = 1-x+x^2-x^3+\cdots$$ and $$h(x) = 1-2x+3x^2-4x^3+\cdots$$, and to realize that we’re asking for $$g(1)$$ and $$h(1)$$, respectively. The sum defining $$g(x)$$ converges for $$-1 < x < 1[/latex], and since $g(x) + x \cdot g(x) = 1$ (this is just the "trick" from above!), we find $g(x) = 1/(1+x)$. So, even though the sum defining $g(1) = 1-1+1-\cdots$ does not converge, the limit as $x$ approaches 1 from below is well defined and equals $1/(1+1) = 1/2$. Likewise, we find that $h(x) = 1/(1+x)^2$ for $-1 < x < 1$, and the limit as $x$ goes to 1 is 1/4. This method of putting the terms as the coefficients of a power series and extrapolating with limits is called Abel Regularization. Notice, however, that this method does not work for the sum [latex]Z$$, because $$1 + 2x + 3x^2 + 4x^3 + \cdots = 1/(1-x)^2$$, and the limit as $$x$$ approaches 1 is still infinite.

A different method of summing divergent series is called Zeta Function Regularization: to compute $$a_1 + a_2 + a_3 + \cdots$$, define the function $$a_1^{-s} + a_2^{-s} + a_3^{-s} + \cdots$$ (which hopefully converges for large enough s), and try to “extend” this to a function we can evaluate at $$s=-1$$. For example, for $$H = 1-2+3-\cdots$$, the corresponding function $$1/1^2-1/2^s+1/3^s-\cdots$$ is called the Dirichlet Eta Function, $$\eta(s)$$. Analysis based on our “trick” from above (iterated infinitely many times! [producing the so-called Euler Transform]) can be used to show that $$\eta(s)$$ can be defined for all values $$s$$ (even complex values!) and that $$\eta(-1) = 1/4$$, which agrees with our computation of $$1-2+3-4+\cdots$$ above. The corresponding function for the sum $$Z$$ is called the Riemann Zeta Function, $$\zeta(s)$$, and we can compute (for all $$s$$ where this makes sense): $$\begin{gather*} \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots,\\ 2^{-s} \zeta(s) = \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \frac{1}{8^s} + \cdots, \text{ and so}\\ \zeta(s)-2\cdot 2^{-s} \zeta(s) = \frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots = \eta(s).\end{gather*}$$ So $$(1-2^{1-s}) \zeta(s)$$ equals $$\eta(s)$$ everywhere (by general complex analysis facts). At $$s = -1$$, this says $$\zeta(-1) = \eta(-1)/(-3) = -1/12$$, as claimed.