Tag Archives: eta function

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.