Are there any real analytic functions such that $\sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{k=2}^{\infty} f(k) \left( \zeta(k) - 1 \right) $?

Question

I wonder whether there are any real functions that are analytic on $\mathbb{R}_{\geq 1}$. that satisfy the equation $$\sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{k=2}^{\infty} f(k) \left( \zeta(k) - 1 \right). \tag{1} $$

Here, $\zeta(\cdot)$ is the Riemann zeta function. If $f(k) \in \mathbb{Q}$ for all $k>1$, the sum on the right is a rational zeta series.

The function $f(n) = n^2$ is a (somewhat) near miss, as we have $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2) \tag{2}, $$ and $$\sum_{k=2}^{\infty} k^{2} \left( \zeta(k) -1 \right) = 1 + \frac{\zeta(2)}{3} + 2 \zeta(3). \tag{3} $$

Questions:

1. Are there any functions $f(\cdot)$ that are real analytic on $\mathbb{R}_{\geq 1}$ such that they satisfy equation $(1)$ ?
2. Can we characterize all such solutions such that $f(k) \in \mathbb{Q}$ for all $k\geq1$ ?

Answer 1

$1$

Yes! There are different $f\left( x \right)$ that are analytical for $x \in \mathbb{R}_{\geq 1}$. However, I personally can't find a closed form for any of these functions. However, we can find them using Newton's method. We can do this with many functions.

$1.1$ Iff $f\left( x \right) = x^{b}$

While playing around with Desmos, I noticed that for $f\left( x \right) = x^{b}$ there seems to be a value $b$ for which $\sum\limits_{n = 1}^{\infty}\left[ \frac{1}{f\left( n \right)} \right] = \sum\limits_{k = 2}^{\infty}\left[ f\left( k \right) \cdot \left( \zeta\left( k \right) - 1 \right) \right]$ applies because when sampling the interval $b \in \left[ 1,\, 2 \right]$ a sign change of $\lim\limits_{a \to \infty}\left[ \sum\limits_{n = 1}^{a}\left[ \frac{1}{f\left( n \right)} \right] - \sum\limits_{k = 2}^{a}\left[ f\left( k \right) \cdot \left( \zeta\left( k \right) - 1 \right) \right] \right] = \zeta\left( b \right) - \sum\limits_{k = 2}^{\infty}\left[ f\left( k \right) \cdot \left( \zeta\left( k \right) - 1 \right) \right] =: g\left( b \right)$ occurs.

Solving for $b$ would be difficult algebraically (except for $a = 2$ $\implies$ $b = \frac{\ln\left(\frac{\sqrt{6 \cdot \pi^{2} - 27} + 3}{\pi^{2} - 6} \right)}{\ln\left( 2 \right)} \approx 1,1649$), but solving numerically would be possible.

Let's say $b_{\infty}$ is a zero of $g\left( x \right)$, then we can find $b_{\infty}$ using Newton's methods: $$ \begin{align*} b_{n + 1} &= b_{n} - \frac{g\left( b_{n} \right)}{g'\left( b_{n} \right)}\\ b_{n + 1} &= b_{n} - \frac{\zeta\left( b_{n} \right) - \sum\limits_{k = 2}^{a}\left[ k^{b_{n}} \cdot \left( \zeta\left( k \right) - 1 \right) \right]}{\zeta'\left( b_{n} \right) - \sum\limits_{k = 2}^{a}\left[ \ln\left( k \right) \cdot k^{b_{n}} \cdot \left( \zeta\left( k \right) - 1 \right) \right]}\\ \end{align*} $$

If $b_{0} = 2$ then $b \equiv b_{\infty} = 1.3193189491950006...$.

So: $$ \fbox{$ \begin{align*} f\left( x \right) &= x^{1.3193189491950006...}\\ \end{align*} $} $$

($g\left( 1.3193189491950006 \right) \approx -1.29763 \cdot 10^{-12}$)

We can also specify a $b$ using series expansion (it would look like this: $b = 2 - \sum\limits_{k = 1}^{\infty}\left[ \operatorname{sgn}\left( c_{k} \right) \cdot \left(\frac{1}{2} \right)^{k} \right]$ with $c_{k} = g\left( 2 - \sum\limits_{m = 1}^{k - 1}\left[ \operatorname{sgn}\left( c_{m} \right) \cdot \left(\frac{1}{2} \right)^{m} \right] \right)$), but Wolfram|Cloud had a few problems computing this, So I rejected that idea, but maybe someone else will do it.

This is a plot of $g\left( b \right)$ by Wolfram|Cloud:

$1.2$ Iff $f\left( x \right) = a^{x}$

Here we find the same thing as with $1.1$, and we can do exactly the same thing again here. Let's say $j\left( a \right) := \sum\limits_{n = 1}^{\infty}\left[ \frac{1}{f\left( n \right)} \right] - \sum\limits_{k = 2}^{\infty}\left[ f\left( k \right) \cdot \left( \zeta\left( k \right) - 1 \right) \right] = \sum\limits_{n = 1}^{\infty}\left[ a^{-k} \right] - \sum\limits_{k = 2}^{\infty}\left[ a^{k} \cdot \left( \zeta\left( k \right) - 1 \right) \right]$. However, Wolfram|Cloud has problems with the expression, so let's simplify it: If $\left| a \right| \leq 1$ then the first term dierges, so $\left| a \right| > 0$ and with this we get $j\left( a \right) = \frac{1}{a - 1} - \left( \sum\limits_{k = 2}^{\infty}\left[ a^{k} \cdot \zeta\left( k \right) \right] + \frac{a^{2}}{a - 1} \right)$. Iff $1 < a < 2$ (diverges for $a > 2$ to $-\infty$) then $\sum\limits_{k = 2}^{\infty}\left[ a^{k} \cdot \zeta\left( k \right) \right] = -a \cdot \left( \gamma + \psi^{\left( 0 \right)}\left( 1 - a \right) \right)$, so $j\left( a \right) \equiv \frac{1 - a \cdot \left( a + \left( 1 - a \right) \cdot \left( \gamma + \psi^{\left( 0 \right)}\left( 1 - a \right) \right) \right)}{a - 1}$ where $\psi^{0}$ is the polygamma functions ($\psi^{\left( 0 \right)}\left( x \right) \equiv \frac{\operatorname{d}}{\operatorname{d}x} \ln\left( \Gamma\left( x \right) \right)$). Wolfram|Alpha simplifies it even further to $j\left( a \right) = -1 + a \cdot \left( \gamma - 1 + \psi^{\left( 0 \right)}\left( 1 - a \right) \right)$. With $a_{0}$ we get: $$ \begin{align*} f\left( x \right) &= 1.3814583044735516...^{x}\\ \end{align*} $$

($j\left( 1.3814583044735516 \right) \approx 4.440892098500626 \cdot 10^{-16}$)

A plot of $j\left( a \right)$ by Wolfram|Cloud:

$1.3$ Table of Values

$n$	$b_{n}$ (rounded)	$a_{n}$ (rounded)
$0$	$2$	$1.5$
$1$	$1.4148332956$	$1.3854522384958676$
$2$	$1.3059582358238548$	$1.3814491248358245$
$3$	$1.3189441203442636$	$1.3814583044222533$
$4$	$1.3193186648698678$	$1.3814583044735516$
$5$	$1.3193189491948372$	$1.3814583044735516$
$6$	$1.3193189491950008$	$1.3814583044735516$
$7$	$1.3193189491950006$	$1.3814583044735516$
$8$	$1.3193189491950006$	$1.3814583044735516$