There is a function in Riemann's 1859 paper (I will use $J(x)$ as in Edward's book), defined as \begin{equation*} J(x) = \frac{1}{2}\left[ \sum_{p^n < x} \frac{1}{n} + \sum_{p^n \leq x} \frac{1}{n} \right]\quad\quad x > 0. \end{equation*} So, for any given $x > 0$, $J(x)$ is a sum over all whole numbers $\leq x$. For each such number, that term in the sum is $1$ for every prime, $1/2$ for every prime square, $1/3$ for every prime cube, etc., and $0$ if that number is a factor of two or more different primes. (For our purposes, it is not important that the step is only half the stated value if $x$ exactly equals a number factored by only one prime).
It is clear that $J(x) < x$ because no term is greater than $1$ and some terms are less than $1$. But there is another definition of $J(x)$ in Riemann's paper \begin{equation} J(x) = \sum_{n=1}^{\infty} \frac{1}{n} \pi\left(x^{1/n}\right) = \pi(x) + \frac{1}{2} \pi(x^{1/2}) + \frac{1}{3} \pi(x^{1/3}) + ..., \end{equation} where $\pi(x)$ has its usual definition as the prime counting function.
I have attempted (unsuccessfully) to prove that $J(x) < x$ using the second definition of $J(x)$. I will show what I have done so far.
We can assume that $x \geq 2$ because $\pi(x)=0$ (and therefore $J(x) = 0$) for $x < 2$. Fix $x$. Then there is a $K \geq 1$ such that $x = 2^K$, and there is a whole number $N$ such that $N \leq K < (N+1)$. Note that for whole number $n > N$ we have $x^{1/n} < 2$ and therefore $\pi(x^{1/n}) = 0$. Thus, for our fixed $x$ (and for our associated $N$ and $K$) we have \begin{equation*} J(x) = \sum_{n=1}^{N} \frac{1}{n} \pi\left(x^{1/n}\right) . \end{equation*} Now assume that $x > 8$. (With $x \leq 8$ there are at most three non-zero terms in $J(x)$ and $J(x) < x$ is quickly verified by hand). For $x > 8$, it is easily seen that $\pi(x) < x/2$ since: (1) there are no primes in the 50% of remaining numbers that are factors of two, and (2) there are no primes in the 16.67% of remaining numbers that are odd factors of three.
Replacing $\pi(x)$ with $x/2$, we have \begin{equation*} J(x) \leq \sum_{n=1}^N \frac{1}{n}\cdot \frac{x^{1/n}}{2} = \sum_{n=1}^N \frac{x^{1/n}}{2n} = \frac{1}{2} \sum_{n=1}^N \frac{x^{1/n}}{n}. \end{equation*} The Inequalities. So we would like to show that \begin{equation} \frac{1}{2} \sum_{n=1}^N \frac{x^{1/n}}{n} < x\quad\text{(or equally)}\quad \frac{1}{2} \sum_{n=1}^N \frac{2^{K/n}}{n} < 2^K. \end{equation} I have satisfied myself, empirically, that the inequalities hold (i.e., even after our rough estimate for $\pi(x)$). But if there is an arithmetico-geometric-like closed form expression (equality or useful inequality), or other way to proceed, I have been unable to find it.
I'll just note one other way to state the inequalities \begin{equation*} \frac{1}{2} \sum_{n=1}^N \frac{x^{1/n}}{n} = \frac{1}{2} \sum_{n=1}^N \frac{ x \cdot x^{-(n-1)/n} }{n} = x \sum_{n=1}^N \frac{1}{2n x^{(n-1)/n}}, \end{equation*} so an equivalent inequality (using $x$ or $2^K$) is \begin{equation} \sum_{n=1}^N \frac{1}{2n x^{(n-1)/n}} < 1. \end{equation}
My question: How do you prove the above inequalities? If that is not possible, are there other ways to use the second definition of $J(x)$ to show $J(x) < x$ (of course, not including proof that the second definition of $J(x)$ equals the first)?
$N \le \log_{2}x$ so for $x>8, \sum_{n=2}^N \frac{x^{1/n}}{n} \le \frac{1}{2}\sqrt{x}\log_{2}x <x$ Done!
(as $\sqrt{x} - \frac{1}{2\ln 2}\ln x$ has positive derivative when $(\ln 2) (\sqrt{x}) >1$ and that happens already for $x>4$)