Here is a small puzzle. Why is

$ \sum_{\substack{m,n\geq1\cr (m,n)=1}}\frac{1}{(m^2+1)(n^2+1)}=1\,? $

In fact, a parameter gives the slightly surprising identity

$ \boxed{\displaystyle \sum_{\substack{m,n\geq1\cr (m,n)=1}}\frac{1}{(m^2-r)(n^2-r)}=\frac{5-3r}{2(1-r)^2}.} \qquad\text{(1)} $

It holds for every real $r<1$, and indeed for every complex $r$ which is not a positive integer square. The summands have poles at all integer squares, but the answer has a pole only at $r=1$. Where did all the others go?

We give two proofs: one by Euler’s convolution of even zeta values, the other by telescoping on coprime pairs. The latter is a generating-function version of a construction of Don Zagier.

First proof: Euler’s convolution

For $\Re s,\Re t>1$, separating the greatest common divisor in the product $\zeta(s)\zeta(t)$ gives

$ \sum_{\substack{m,n\geq1\cr (m,n)=1}}\frac{1}{m^s n^t}=\frac{\zeta(s)\zeta(t)}{\zeta(s+t)}. \qquad\text{(2)} $

Write $S(r)$ for the left side of (1). For $\lvert r\rvert<1$, geometric expansion and (2) give

$ S(r)=\sum_{h\geq0}r^h\sum_{j=0}^{h}\frac{\zeta(2j+2)\zeta(2h-2j+2)}{\zeta(2h+4)}. $

Now Euler’s convolution formula says that

$ \sum_{j=1}^{N-1}\zeta(2j)\zeta(2N-2j)=\left(N+\frac12\right)\zeta(2N)\qquad(N\geq2). \qquad\text{(3)} $

With $N=h+2$, this turns the double sum into

$ S(r)=\sum_{h\geq0}\left(h+\frac52\right)r^h=\frac{5-3r}{2(1-r)^2}. $

Both sides are holomorphic away from the positive integer squares. The identity theorem proves (1) throughout that domain. We shall prove (3) below, following Zagier.

Second proof: a Stern–Brocot telescoping

The rational function

$ H_r(m,n)=\frac{2m^2+2n^2+mn-3r}{2mn(m^2-r)(n^2-r)} \qquad\text{(4)} $

has the remarkable property

$ \boxed{\displaystyle H_r(m,n)-H_r(m+n,n)-H_r(m,m+n)=\frac{1}{(m^2-r)(n^2-r)}.} \qquad\text{(5)} $

For a quick verification, split $H_r=T_r+I_r$, where

$ T_r(m,n)=\frac{1}{mn}\left(\frac{1}{m^2-r}+\frac{1}{n^2-r}\right),\qquad I_r(m,n)=\frac{mn+r}{2mn(m^2-r)(n^2-r)}. $

Put $u=m+n$ and $\Delta F(m,n)=F(m,n)-F(u,n)-F(m,u)$. Then

$ \Delta T_r=\frac{m^2+mn+n^2-3r}{(m^2-r)(n^2-r)(u^2-r)},\qquad \Delta I_r=\frac{mn+2r}{(m^2-r)(n^2-r)(u^2-r)}. $

Their sum is $1/((m^2-r)(n^2-r))$, as claimed.

Every coprime pair other than $(1,1)$ has exactly one parent, obtained by subtracting the smaller coordinate from the larger. Thus the two maps

$ (m,n)\longmapsto(m+n,n),\qquad (m,n)\longmapsto(m,m+n) $

enumerate all coprime positive pairs except $(1,1)$. Summing (5) ought to leave only $H_r(1,1)$, which is precisely the right side of (1).

Here is the small convergence detail. Let $Q_N$ be the coprime pairs with $1\leq m,n\leq N$. Cancellation gives

$ \sum_{(m,n)\in Q_N}\Delta H_r(m,n)=H_r(1,1)-\sum_{\substack{(m,n)\in Q_N\cr m+n>N}}\bigl(H_r(m+n,n)+H_r(m,m+n)\bigr). $

For fixed $r$ away from the positive squares,

$ \lvert H_r(x,y)\rvert\ll_r\frac{1}{xy^3}+\frac{1}{x^2y^2}+\frac{1}{x^3y}\qquad(x,y\geq1). $

For each $n\leq N$, at most $n$ values of $m\leq N$ have $m+n>N$. The displayed bound makes the entire boundary sum $O_r(N^{-1})$. The summand in (1) is absolutely summable, so (5) telescopes rigorously to $S(r)=H_r(1,1)$.

Zagier’s identity

Zagier [1994, pp. 498–499] starts with the rational functions, for even $k\geq4$,

$ f_k(m,n)=\frac{1}{mn^{k-1}}+\frac12\sum_{j=2}^{k-2}\frac{1}{m^j n^{k-j}}+\frac{1}{m^{k-1}n}. \qquad\text{(6)} $

His key observation is

$ \boxed{\displaystyle \Delta f_k(m,n)=\sum_{\substack{2\leq j\leq k-2\cr j\ \mathrm{even}}}\frac{1}{m^j n^{k-j}}.} \qquad\text{(7)} $

In fact, our rational function (4) is nothing but the generating function of all these $f_k$:

$ \boxed{\displaystyle H_r(m,n)=\sum_{h\geq0}r^h f_{2h+4}(m,n).} \qquad\text{(8)} $

To see this, sum the two endpoint terms of (6) geometrically; they give $T_r$. The middle terms give $I_r$, by taking the even-degree part of a product of two geometric series. Explicitly,

$ \frac12\sum_{h\geq0}r^h\sum_{j=2}^{2h+2}\frac{1}{m^j n^{2h+4-j}}=\frac{mn+r}{2mn(m^2-r)(n^2-r)}. $

Apply $\Delta$ to (8), use (5), and compare coefficients of $r^h$ in

$ \frac{1}{(m^2-r)(n^2-r)}=\sum_{h\geq0}r^h\sum_{j=0}^{h}\frac{1}{m^{2j+2}n^{2h-2j+2}}. $

This proves (7), including the algebraic verification omitted in Zagier’s short account (and supplied by a different argument in Choi [2009]).

To obtain Euler’s formula, sum (7) over all positive $m,n$. The two child maps now cover all pairs off the diagonal, so that

$ \sum_{m,n\geq1}\Delta f_k(m,n)=\sum_{n\geq1}f_k(n,n)=\frac{k+1}{2}\zeta(k). $

As above, take finite square cutoffs first: $f_k$ satisfies the same type of bound as $H_r$, and the boundary error tends to zero. The right side of (7), summed over all pairs, is $\sum_{2\leq j\leq k-2,\ j\ \mathrm{even}}\zeta(j)\zeta(k-j)$. Therefore

$ \boxed{\displaystyle \sum_{\substack{2\leq j\leq k-2\cr j\ \mathrm{even}}}\zeta(j)\zeta(k-j)=\frac{k+1}{2}\zeta(k)\qquad(k\geq4\ \mathrm{even}).} $

This is Euler’s formula (3). Zagier’s summation over all pairs leaves the diagonal; summation over coprime pairs leaves just its first point $(1,1)$.

Where have the poles gone?

There is an unexpectedly simple answer. For every integer $k\geq2$,

$ \boxed{\displaystyle \sum_{\substack{n\geq1\cr (n,k)=1}}\frac{1}{n^2-k^2}=0.} \qquad\text{(9)} $

Indeed,

$ \frac{1}{n^2-k^2}=\frac{1}{2k}\left(\frac{1}{n-k}-\frac{1}{n+k}\right). $

Since coprimality with $k$ is unchanged by translating an integer by $k$, shifting the two sums (first with a finite cutoff) reduces their difference to

$ \frac{1}{2k}\sum_{\substack{-k<t<k\cr (t,k)=1}}\frac1t=0, $

by pairing $t$ with $-t$; the cutoff error is $O_k(N^{-1})$. The residue of the double sum at $r=k^2$ is $-2$ times the left side of (9), so it vanishes. Thus the apparent singularities at $4,9,16,\ldots$ are removable. The original summands are undefined there, of course, but the sum has a finite analytic continuation.

Two more consequences are worth recording. By symmetry, the diagonal contributes only $(m,n)=(1,1)$, and so (1) gives the even simpler identity

$ \boxed{\displaystyle \sum_{\substack{1\leq m<n\cr (m,n)=1}}\frac{1}{(m^2-r)(n^2-r)}=\frac{3}{4(1-r)}.} \qquad\text{(10)} $

Also, restricting to pairs of fixed greatest common divisor $d$, rather than to coprime pairs, amounts to replacing $(m,n)$ by $(dm,dn)$:

$ \sum_{\substack{m,n\geq1\cr (m,n)=d}}\frac{1}{(m^2-r)(n^2-r)}=\frac{5d^2-3r}{2d^2(d^2-r)^2}. \qquad\text{(11)} $

The cotangent behind the sum

Put $A(r)=\sum_{n\geq1}(n^2-r)^{-1}$. The classical partial-fraction expansion of the cotangent is

$ \pi\cot(\pi z)=\frac1z+\sum_{n\geq1}\frac{2z}{z^2-n^2}=\frac1z-2zA(z^2). $

Now $p(z)=\pi\cot(\pi z)$ satisfies $p’(z)=-\pi^2-p(z)^2$. Substituting $p(z)=z^{-1}-2zA(z^2)$, with $r=z^2$, we obtain

$ -\frac1{z^2}-2A(r)-4rA’(r)=-\pi^2-\left(\frac1z-2zA(r)\right)^2, $

or simply

$ \boxed{\displaystyle A(r)^2=A’(r)+\frac{3}{2r}\bigl(A(r)-\zeta(2)\bigr).} \qquad\text{(12)} $

But (12) is also what we get by forgetting coprimality. Indeed, grouping all positive pairs by their greatest common divisor and using (1),

$ \begin{aligned} A(r)^2=\sum_{d\geq1}\frac1{d^4}S\left(\frac{r}{d^2}\right)=\sum_{d\geq1}\frac{5d^2-3r}{2d^2(d^2-r)^2}=\sum_{d\geq1}\left[\frac1{(d^2-r)^2}+\frac{3}{2r}\left(\frac1{d^2-r}-\frac1{d^2}\right)\right],\end{aligned} $

which is exactly (12). Conversely, Mobius inversion gives $S(r)=\sum_{d\geq1}\mu(d)d^{-4}A(r/d^2)^2$. Thus (1) is the coprime version of the cotangent Riccati equation.

Finally, expanding $A(r)=\sum_{j\geq0}\zeta(2j+2)r^j$ and comparing coefficients in (12) gives

$ \sum_{j=0}^h\zeta(2j+2)\zeta(2h-2j+2)=\left(h+\frac52\right)\zeta(2h+4), $

which is just Euler’s formula (3).

A natural question is what happens with two different parameters $r$ and $s$. Then the coefficient of $r^i s^j$ is

$ \frac{\zeta(2i+2)\zeta(2j+2)}{\zeta(2i+2j+4)}, $

and it is only on the diagonal $r=s$ that Euler’s convolution immediately collapses all coefficients of a given total degree.

References

  1. L. Euler. The convolution recurrence for even zeta values; see DLMF 25.6.16. The cotangent partial-fraction formula is DLMF 4.22.3.
  2. D. Zagier. Values of Zeta Functions and Their Applications, First European Congress of Mathematics, Vol. II, Progress in Mathematics 120, Birkhauser, 1994, 497–512, especially pp. 498–499.
  3. J. Choi. Several Results Associated with the Riemann Zeta Function, Journal of the Chungcheong Mathematical Society 22 (2009), 467–480, Section 2.