# Monthly Archives: November 2018

## The wrong way to compute a sum: addendum

In my previous note, I looked at an amusing but inefficient way to compute the sum $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1}$$ using Mellin and inverse Mellin transforms. This was great fun, but the amount of work required was more intense than the more straightforward approach offered immediately by using Lambert series.

However, Adam Harper suggested that there is a nice shortcut that we can use (although coming up with this shortcut requires either a lot of familiarity with Mellin transforms or knowledge of the answer).

In the Lambert series approach, one shows quickly that $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ and then evaluates this last sum directly. For the Mellin transform approach, we might ask: do the two functions $$f(x) = \sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1}$$ and $$g(x) = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ have the same Mellin transforms? From the previous note, we know that they have the same values at $1$.

We also showed very quickly that $$\mathcal{M} [f] = \frac{1}{(\log 2)^2} \Gamma(s) \zeta(s-1).$$ The more difficult parts from the previous note arose in the evaluation of the inverse Mellin transform at $x=1$.

Let us compute the Mellin transform of $g$. We find that \begin{align} \mathcal{M}[g] &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{2^{nx}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{e^{nx \log 2}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} \frac{n}{(n \log 2)^s} \int_0^\infty x^s e^{-x} \frac{dx}{x} \notag \\ &= \frac{1}{(\log 2)^2} \zeta(s-1)\Gamma(s). \notag \end{align} To go from the second line to the third line, we did the change of variables $x \mapsto x/(n \log 2)$, yielding an integral which is precisely the definition of the Gamma function.

Thus we see that $$\mathcal{M}[g] = \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1) = \mathcal{M}[f],$$ and thus $f(x) = g(x)$. (“Nice” functions with the same “nice” Mellin transforms are also the same, exactly as with Fourier transforms).

This shows that not only is $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ but in fact $$\sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1} = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ for all $x > 1$.

I think that’s sort of slick.

## The wrong way to compute a sum

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 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

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.