Proof of the Fractional Sum Definition of the Zeta Function

This post contains a proof of the contents of my video, Extending the Zeta Function with Fractional Sums. If you aren’t coming from that video, I highly recommend watching it first.

My new video builds on a previous video in which I arrive at a definition for fractional sums:

$$\sum_{k=1}^z f(k) = \lim_{n \to \infty} \left( \sum_{k=1}^{n-1} (f(k) - f(z + k)) + \sum_{k=1}^m \binom zk \Delta^{k-1} f(n) \right). \tag{1}$$

Here $m$ can be any natural number and $f$ must be a sufficiently well-behaved function that grows strictly slower than $z^m$. In my new video I use these fractional sums to derive a formula for the Riemann zeta function:

$$\zeta(s) = \frac1{s-1} \frac d{dz}\left. \sum_{k=1}^z k^{1-s} \right|_{z=0} \qquad \text{Re}(s) > 1 - m, \tag{2}$$

for whatever $m$ we choose in the $(1)$. However, I did not prove that the fractional sum in this formula converges, that a necessary interchange of limits is valid, or that this new definition of $\zeta$ is holomorphic for all $s$ in its domain.

This post contains a proof of these facts.

Locally Uniform Convergence is Enough

In my video, we kicked things off by asking what the fractional sum of $z \mapsto z^w$ looks like for any fixed $w \in \mathbb{C}$. We used the principal branch of this function such that $1^w = 1$, with a branch cut on $(-\infty, 0]$. It then follows from $(1)$ that the fractional sum

$$z \mapsto \sum_{k=1}^z k^w$$

has a branch cut on $(-\infty, -1]$.

We then observed that this fractional sum appeared to be holomorphic in $\mathbb{C} \setminus (-\infty, 1]$ as long as the real part of $w$ is less than our choice of $m$ in $(1)$. For the rest of this post we will assume a fixed $m$, and we will examine the fractional sum as a function of both $z$ and $w$.

Definitions: The domain $\Omega$ is given by

$$\Omega = (\mathbb{C} \setminus (-\infty, 1]) \times \{w \in \mathbb{C} : \text{Re}(w) < m\}.$$

For each $n \in \mathbb{N}$, define $F_n: \Omega \to \mathbb{C}$ by

$$F_n(z, w) = \sum_{k=1}^{n-1} (k^w - (z + k)^w) + \sum_{k=1}^m \binom zk \Delta_n^{k-1} (n^w).$$

Define $F: \Omega \to \mathbb{C}$ as the limit

$$F(z, w) = \lim_{n \to \infty} F_n(z, w)$$

if the sequence $F_n$ converges.

This function $F$ is the fractional sum of $z^w$.

To show that $(2)$ is the analytic continuation of the zeta function, we must prove that when $w$ is fixed, $z \mapsto F(z, w)$ is holomorphic so that we may take its derivative at $z=0$. We also need that $\frac{\partial F_n}{\partial z} \to \frac{\partial F}{\partial z}$ as $n \to \infty$ because if we can swap the derivative and the limit, we can get

$$\left.\frac{\partial F}{\partial z}(z, 1-s)\right|_{z=0} = \lim_{n \to \infty} \left(\sum_{k=1}^{n-1} \frac1{k^s} + \sum_{k=1}^m \frac{(-1)^k \Delta^{k-1}_n (n^{1-s})}{k(1 - s)} \right).$$

The second summation vanishes in the limit when $\text{Re}(s) > 1$, so this reduces to usual the infinite series definition of $\zeta(s)$. Therefore, if $w \mapsto \frac{\partial F}{\partial z}(0,w)$ is holomorphic, we have the analytic continuation of $\zeta$.

Fortunately, “holomorphic functions are really nice”. We can simply use a standard complex analysis result which I’ll refer to the uniform convergence theorem:

The Uniform Convergence Theorem: If a sequence of holomorphic functions $f_n$ converges to a function $f$ locally uniformly (or equivalently, uniformly on compact sets), then $f$ is holomorphic, and $f_n'$ converges locally uniformly to $f'$.

So all we need to do is show that $F_n$ converges locally uniformly, and we’ll have all the conditions we need.

The only catch (for me anyway) is that this uniform convergence theorem is only for functions of a single variable. It is not immediately clear that locally uniform convergence allows us to conclude that $\frac{\partial F}{\partial z}$ is holomorphic in $w$ for fixed $z$. I am sure that this is a standard result in multi-variable complex analysis, but for me, it’s a “missing link” in the proof. Fortunately, for our specific sequence $F_n$, we can prove this quickly with Cauchy’s integral formula.

Proof of Locally Uniform Convergence

We start by defining the following domains:

Definition: Given any $R \in \mathbb{N}$, let $\varepsilon = 1/R$ and define

$$\begin{align*} U_R &= \{z \in \mathbb{C} : |z| < R\} \setminus (-\infty, -1],\\ V_R &= \{w \in \mathbb{C} : |w| < R,\ \ \text{Re}(w) < m - \varepsilon\}. \end{align*}$$

For any $(z, w) \in \Omega$, we can choose $R$ to be large enough that $(z, w) \in U_R \times V_R$. Therefore it suffices to prove that $f_n$ converges uniformly in $U_R \times V_R$ for arbitrarily large $R$. For the rest of the proof, we will assume an arbitrary fixed $R > m$, and thus a fixed $\varepsilon = 1/R$. We will use $C$ to denote a positive constant which may change from line to line, but depends only on $m$ and $R$.

We will start by demonstrating that any degree $m$ polynomial is bounded on $U_R$ by its values at the integers $0$ through $m$.

Lemma: Let $p(z)$ be a degree $m$ polynomial. Then there exists a constant $C$ such that for any $z \in U_R$,

$$|p(z)| \leq C \cdot \max_{\substack{k \in \mathbb{Z} \\ 0\leq k\leq m}} |p(k)|.$$

Proof: Consider the Lagrange polynomial representation of $p(z)$ determined by its values at $0, 1, \dots, m$:

$$p(z) = \sum_{k=0}^m p(k) \prod_{\substack{0 \leq j \leq m \\ j \neq k}} \frac{z - j}{k - j}.$$

Since $|z - j| \leq |z| + |j| \leq R + m$ and since $|k - j| \geq 1$, we have

$$\begin{aligned} |p(z)| &\leq \sum_{k=0}^m |p(k)| (R + m)^m \\ &\leq (m+1)(R+m)^m \cdot \max_{\substack{k \in \mathbb{Z} \\ 0\leq k\leq m}} |p(k)|. \end{aligned}$$

Thus the claim is established with $C = (m+1)(R+m)^m$. $\square$

The specific polynomials we will work with are the degree $m$ Gregory-Newton and Taylor polynomials of $(z+n)^w$ centered at $z=0$. Let us introduce shorthands for them.

Definition: Let $n \in \mathbb{N}$. We define

$$\begin{aligned} G_n(z,w) &= \sum_{k=0}^m \binom zk \Delta_n^k(n^w),\\ T_n(z,w) &= \sum_{k=0}^m \frac{z^k}{k!}\!\left.\frac{d^k}{d\zeta^k}(\zeta+n)^w\right|_{\zeta=0}. \end{aligned}$$

We can now begin our proof that $F_n$ converges uniformly on $U_R \times V_R$.

Theorem: $F_n$ converges uniformly on $U_R \times V_R$.

Proof: Observe that if $N > R$, then

$$F_N(z,w) = F_{R+1}(z,w) + \sum_{n=R+1}^N \Delta_n F_n(z,w), \tag{3}$$

so it suffices to establish uniform convergence of the series

$$\sum_{n=R+1}^\infty \Delta_n F_n(z,w).$$

Let $(z, w) \in U_R \times V_R$. By inserting the definition of $F_n$ and grouping terms, we notice that

$$\begin{aligned} \Delta_n F_n(z,w) &= (n^w - (z+n)^w) + \Delta_n\sum_{k=1}^m \binom zk \Delta_n^{k-1} (n^w)\\ &= n^w + \sum_{k=1}^m \binom zk \Delta_n^k (n^w) - (z+n)^w\\ &= \sum_{k=0}^m \binom zk \Delta_n^k (n^w) - (z+n)^w\\ &= G_n(z,w) - (z+n)^w. \end{aligned}$$

Therefore

$$\begin{align} |\Delta_n F_n(z,w)| &= |G_n(z,w) - (z+n)^w| \notag\\ &\leq |G_n(z,w) - T_n(z,w)| + |T_n(z,w) - (z+n)^w|. \tag{4} \end{align}$$

Since $U_R$ contains the line segment from $0$ to $z$, Taylor’s theorem with remainder gives

$$\begin{aligned} |T_n(z,w) - (z+n)^w| &\leq \frac{|z|^{m+1}}{(m+1)!} \cdot \sup_{\zeta \in U_R}\left|\frac{d^{m+1}}{d\zeta^{m+1}} (\zeta+n)^w\right|\\ &< C \cdot \sup_{\zeta \in U_R}\left|\frac{d^{m+1}}{d\zeta^{m+1}} (\zeta+n)^w\right|. \end{aligned}$$

By differentiating explicitly and using that $w \in V_R$, hence $|w| < R$, we obtain

$$\begin{aligned} \left|\frac{d^{m+1}}{d\zeta^{m+1}} (\zeta+n)^w\right| &= |w||w-1|\cdots|w-m|\left| (\zeta+n)^{w-m-1} \right|\\ &\leq C \cdot |\zeta + n|^{\text{Re}(w) - m - 1}. \end{aligned}$$

Since $\text{Re}(w) < m - \varepsilon$, we see that $\text{Re}(w) - m - 1 < -(1 + \varepsilon)$. Also, since $|\zeta| < R \leq n - 1$, we can see that $|\zeta + n| \geq n - |\zeta| > 1$. Therefore

$$\begin{aligned} |\zeta + n|^{\text{Re}(w) - m - 1} &\leq (n - |\zeta|)^{-(1 + \varepsilon)}\\ & < (n - R)^{-(1 + \varepsilon)}. \end{aligned}$$

This means that

$$|T_n(z,w) - (z+n)^w| < C (n - R)^{-(1 + \varepsilon)} \tag{5}$$

for some constant $C$. This is a uniform bound on the second part of $(4)$. Let’s turn our attention to the remaining part.

For fixed $w$, the expression $G_n(z,w) - T_n(z,w)$ is a polynomial of degree $m$ in $z$, so our lemma guarantees that

$$|G_n(z,w) - T_n(z,w)| \leq C \cdot \max_{\substack{k \in \mathbb{Z} \\ 0\leq k\leq m}} |G_n(k,w) - T_n(k,w)|.$$

Because the Gregory-Newton polynomial coincides with the function it interpolates at the integers $0$ through $m$, we have

$$|G_n(k,w) - T_n(k,w)| = |(k+n)^w - T_n(k,w)|.$$

Because we chose $R$ to be greater than $m$, we know that $k \in U_R$, and therefore we may apply $(5)$ with $k$ in place of $z$ to see that

$$|G_n(k,w) - T_n(k,w)| \leq C (n - R)^{-(1 + \varepsilon)}$$

for some constant $C$.

We have shown that both parts of the right-hand side of $(4)$ are bounded by $C(n-R)^{-(1+\varepsilon)}$. Therefore, since

$$\sum_{n=R+1}^\infty (n-R)^{-(1+\varepsilon)} = \sum_{n=1}^\infty \frac1{n^{1+\varepsilon}} < \infty,$$

we know by the Weierstrass M-test that

$$\sum_{n=R+1}^\infty \Delta_n F_n(z, w)$$

converges uniformly. Thus $(3)$ shows that $F_N$ converges uniformly on $U_R \times V_R$, so we are done.
$\square$

Bonus Content! An Alternative Uniform Bound

The proof of uniform convergence boiled down to finding a uniform bound for $G_n(z, w) - (z + n)^w$. In $(4)$, we split this into $G_n - T_n$ and $T_n - (z+n)^w$, each of which we bounded separately. However, I recently learned of a more direct way to obtain a bound without introducing a Taylor polynomial. This alternative approach uses an amazing identity called the Hermite-Genocchi formula.

This formula expresses a divided difference in terms of an integral over all convex combinations of the data points. Specifically, let $\tau_n$ denote the $n$-dimensional standard simplex - the set of all tuples $(t_1, \dots, t_n)$ of positive numbers whose sum is at most $1$. Given such a tuple, define $t_0 = 1 - t_1 - \cdots - t_n$, so that $t_0 + \cdots + t_n = 1$. The Hermite-Genocchi formula states that

$$f[z_0, \dots, z_n] = \int_{\tau_n} f^{(n)}(t_0z_0 + \cdots + t_nz_n) dt_n \cdots dt_1,$$

as long as $f^{(n)}$ exists and is continuous in the convex hull of $\{z_0, \dots, z_n\}$.

In the proof above, we needed to bound $|(n+z)^w - G_n(z,w)|$. Since (for fixed $w$) $G_n$ is the interpolating polynomial of $(n+z)^w$ determined by the points $0, \dots, m$, the Newton form of the error gives

$$(n+z)^w - G_n(z, w) = f_{n,w}[0, \dots, m, z] \prod_{k=0}^m (z-k), \tag{6}$$

where $f_{n,w}(z) = (n+z)^w$. Since the convex hull of $\{0, \dots, m, z\}$ is contained in $U_R$ whenever $z \in U_R$, we have by the Hermite Genocchi formula

$$\begin{aligned} \left|f_{n,w}[0, \dots, m, z]\right| &\leq \int_{\tau_{m+1}} \left|f_{n,w}^{(m+1)}(1t_1 + \cdots + mt_m + zt_{m+1})\right|dt_{m+1}\cdots dt_1\\ &\leq \int_{\tau_{m+1}} \sup_{\zeta \in U_R} \left|f_{n,w}^{(m+1)}(\zeta)\right|dt_{m+1}\cdots dt_1\\ &= \frac1{(m+1)!} \sup_{\zeta \in U_R} \left|\frac{d^{m+1}}{d\zeta^{m+1}}(\zeta + n)^w\right|. \end{aligned}$$

Thus, by repeating the argument in the proof above, we arrive at

$$\left|f_{n,w}[0, \dots, m, z]\right| < C(n-R)^{-(1+\varepsilon)}.$$

Finally, since the product in $(6)$ is uniformly bounded for $z \in U_R$, we have achieved a uniform bound for $|(n+z)^w - G_n(z, w)|$ – the same bound as in the proof above (up to the constant $C$).