Tag Archives: natural numbers

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 \(-1 < x < 1[/latex], and since [latex]g(x) + x \cdot g(x) = 1[/latex] (this is just the "trick" from above!), we find [latex]g(x) = 1/(1+x)[/latex]. So, even though the sum defining [latex]g(1) = 1-1+1-\cdots[/latex] does not converge, the limit as [latex]x[/latex] approaches 1 from below is well defined and equals [latex]1/(1+1) = 1/2[/latex]. Likewise, we find that [latex]h(x) = 1/(1+x)^2[/latex] for [latex]-1 < x < 1[/latex], and the limit as [latex]x[/latex] goes to 1 is 1/4. This method of putting the terms as the coefficients of a power series and extrapolating with limits is called Abel Regularization. Notice, however, that this method does not work for the sum [latex]Z\), 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.