mixedmath

Explorations in math and programming
David Lowry-Duda

At a recent colloquium at the University of Warwick, the fact that $$\label{question} \sum_ {n \geq 1} \frac{\varphi(n)}{2^n - 1} = 2.$$ Although this was mentioned in passing, John Cremona asked — How do you prove that?

It almost fails a heuristic check, as one can quickly check that $$\label{similar} \sum_ {n \geq 1} \frac{n}{2^n} = 2,$$ which is surprisingly similar to \eqref{question}. I wish I knew more examples of pairs with a similar flavor.

[Edit: Note that an addendum to this note has been added here. In it, we see that there is a way to shortcut the "hard part" of the long computation.]

The right way

Shortly afterwards, Adam Harper and Samir Siksek pointed out that this can be determined from Lambert series, and in fact that Hardy and Wright include a similar exercise in their book. This proof is delightful and short.

The idea is that, by expanding the denominator in power series, one has that $$\sum_{n \geq 1} a(n) \frac{x^n}{1 - x^n} \notag = \sum_ {n \geq 1} a(n) \sum_{m \geq 1} x^{mn} = \sum_ {n \geq 1} \Big( \sum_{d \mid n} a(d) \Big) x^n,$$ where the inner sum is a sum over the divisors of $d$. This all converges beautifully for $\lvert x \rvert < 1$.

Applied to \eqref{question}, we find that $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n - 1} \notag = \sum_ {n \geq 1} \varphi(n) \frac{2^{-n}}{1 - 2^{-n}} = \sum_ {n \geq 1} 2^{-n} \sum_{d \mid n} \varphi(d),$$ and as $$\sum_ {d \mid n} \varphi(d) = n, \notag$$ we see that \eqref{question} can be rewritten as \eqref{similar} after all, and thus both evaluate to $2$.

That's a nice derivation using a series that I hadn't come across before. But that's not what this short note is about. This note is about evaluating \eqref{question} in a different way, arguably the wrong way. But it's a wrong way that works out in a nice way that at least one person1 1and perhaps exactly one person finds appealing.

The wrong way

We will use Mellin inversion — this is essentially Fourier inversion, but in a change of coordinates.

Let $f$ denote the function $$f(x) = \frac{1}{2^x - 1}. \notag$$ Denote by $f^ *$ the Mellin transform of $f$, $$f * (s):= \mathcal{M} [f(x)] (s) := \int_ 0^\infty f(x) x^s \frac{dx}{x} = \frac{1}{(\log 2)^2} \Gamma(s)\zeta(s),\notag$$ where $\Gamma(s)$ and $\zeta(s)$ are the Gamma function and Riemann zeta functions.2 2These are functions near and dear to my heart, so I feel comfort when I see them. But I recognize that others might think that this is an awfully complicated way to start answering this question. And I must say, those people are probably right.

For a general nice function $g(x)$, its Mellin transform satisfies $$\mathcal{M}[f(nx)] (s) = \int_0^\infty g(nx) x^s \frac{dx}{x} = \frac{1}{n^s} \int_0^\infty g(x) x^s \frac{dx}{x} = \frac{1}{n^s} g^ * (s).\notag$$ Further, the Mellin transform is linear. Thus $$\label{mellinbase} \mathcal{M}[\sum_{n \geq 1} \varphi(n) f(nx)] (s) = \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} f^ * (s) = \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} \frac{\Gamma(s) \zeta(s)}{(\log 2)^s}.$$

The Euler phi function $\varphi(n)$ is multiplicative and nice, and its Dirichlet series can be rewritten as $$\sum_{n \geq 1} \frac{\varphi(n)}{n^s} \notag = \frac{\zeta(s-1)}{\zeta(s)}.$$ Thus the Mellin transform in \eqref{mellinbase} can be written as $$\frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1). \notag$$

By the fundamental theorem of Mellin inversion (which is analogous to Fourier inversion, but again in different coordinates), the inverse Mellin transform will return the original function. The inverse Mellin transform of a function $h(s)$ is defined to be $$\mathcal{M}^{-1}[h(s)] (x) \notag := \frac{1}{2\pi i} \int_ {c - i \infty}^{c + i\infty} x^s h(s) ds,$$ where $c$ is taken so that the integral converges beautifully, and the integral is over the vertical line with real part $c$. I'll write $(c)$ as a shorthand for the limits of integration. Thus $$\label{mellininverse} \sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} - 1} = \frac{1}{2\pi i} \int_ {(3)} \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1) x^{-s} ds.$$

We can now describe the end goal: evaluate \eqref{mellininverse} at $x=1$, which will recover the value of the original sum in \eqref{question}.

How can we hope to do that? The idea is to shift the line of integration arbitrarily far to the left, pick up the infinitely many residues guaranteed by Cauchy's residue theorem, and to recognize the infinite sum as a classical series.

The integrand has residues at $s = 2, 0, -2, -4, \ldots$, coming from the zeta function ($s = 2$) and the Gamma function (all the others). Note that there aren't poles at negative odd integers, since the zeta function has zeroes at negative even integers.

Recall, $\zeta(s)$ has residue $1$ at $s = 1$ and $\Gamma(s)$ has residue $(-1)^n/{n!}$ at $s = -n$. Then shifting the line of integration and picking up all the residues reveals that $$\sum_{n \geq 1} \frac{\varphi(n)}{2^{n} - 1} \notag =\frac{1}{\log^2 2} + \zeta(-1) + \frac{\zeta(-3)}{2!} \log^2 2 + \frac{\zeta(-5)}{4!} \log^4 2 + \cdots$$

The zeta function at negative integers has a very well-known relation to the Bernoulli numbers, $$\label{zeta_bern} \zeta(-n) = - \frac{B_ {n+1}}{n+1},$$ where Bernoulli numbers are the coefficients in the expansion $$\label{bern_gen} \frac{t}{1 - e^{-t}} = \sum_{m \geq 0} B_m \frac{t^m}{m!}.$$ Many general proofs for the values of $\zeta(2n)$ use this relation and the functional equation, as well as a computation of the Bernoulli numbers themselves. Another important aspect of Bernoulli numbers that is apparent through \eqref{zeta_bern} is that $B_{2n+1} = 0$ for $n \geq 1$, lining up with the trivial zeroes of the zeta function.

Translating the zeta values into Bernoulli numbers, we find that \eqref{question} is equal to \begin{align} &\frac{1}{\log^2 2} - \frac{B_2}{2} - \frac{B_4}{2! \cdot 4} \log^2 2 - \frac{B_6}{4! \cdot 6} \log^4 2 - \frac{B_8}{6! \cdot 8} \cdots \notag \\ &= -\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!}. \label{recog} \end{align} This last sum is excellent, and can be recognized.

For a general exponential generating series $$F(t) = \sum_{m \geq 0} a(m) \frac{t^m}{m!},\notag$$ we see that $$\frac{d}{dt} \frac{1}{t} F(t) \notag =\sum_{m \geq 0} (m-1) a(m) \frac{t^{m-2}}{m!}.$$ Applying this to the series defining the Bernoulli numbers from \eqref{bern_gen}, we find that $$\frac{d}{dt} \frac{1}{t} \frac{t}{1 - e^{-t}} \notag =- \frac{e^{-t}}{(1 - e^{-t})^2},$$ and also that $$\frac{d}{dt} \frac{1}{t} \frac{t}{1 - e^{-t}} \notag =\sum_{m \geq 0} (m-1) B_m \frac{(t)^{m-2}}{m!}.$$ This is exactly the sum that appears in \eqref{recog}, with $t = \log 2$.

Putting this together, we find that $$\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!} \notag =\frac{e^{-\log 2}}{(1 - e^{-\log 2})^2} = \frac{1/2}{(1/2)^2} = 2.$$ Thus we find that \eqref{question} really is equal to $2$, as we had sought to show.

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