I am trying to confirm Littlewoods's Theorem/criteria for Riemann's hypothesis which is based on Mertens' function, $M(x)$, by developing a formula for $M(x)$ as $x \to \infty$:
Theorem-Littlewood The Riemann hypothesis is equivalent to the statement that for every $\epsilon > 0$ the function $M(x)/x^{1/2 + \epsilon}$ approaches zero as $x \to \infty$.
The formula is $$ |M(x)| \simeq \pi(x) \sum_{n=1}^{\infty}\frac{\mu(n)}{n} \qquad \text{as $x \to \infty$}. $$
A result proved by von Mangoldt and de la Vallee Poussin shows that the sum $\sum_{n=1}^{\infty}\frac{\mu(n)}{n}$ equals zero.
Question: Can one then say this implies $|M(x)| \simeq \pi(x)(0)\simeq 0$? Or must one be more careful because $\pi(x$) is approaching $\infty$ so the relation could be indeterminate? [After further study and research my view is that the answer is "One needs to be careful especially in dealing with a limit of the form $\infty \times 0$ or $\infty/\infty$." But this is an old problem in calculus that can be addressed by applying L'Hospital's rule. The scary result however is the analysis implies Riemann's hypothesis is not true!??? See new edits below.]
Question: Would it be more correct (better) to use the following line of reasoning? Divide both sides of the formula by $x^{1/2 + \epsilon}$ which gives $M(x)/x^{1/2 + \epsilon} \simeq \pi(x)/x^{1/2 + \epsilon} \sum_{n=1}^{\infty}\frac{\mu(n)}{n}$ and argue that this equals zero? [My own view now is this is a clearer and more correct approach.]
Since $\pi(x) \simeq x/ \log(x)$ this last expression can be written as $$ \frac{ M(x) }{ x^{1/2 + \epsilon} } \simeq \frac{x / \log(x)}{x^{1/2 + \epsilon}} \sum_{n=1}^{\infty} \frac{\mu(n)}{n} \simeq \frac{x^{1/2 - \epsilon}}{\log(x)} \sum_{n=1}^{\infty}\frac{\mu(n)}{n}. $$
For $\epsilon \geq 1/2$ it seems easy to argue that the right side approaches zero, since if $\epsilon = 1/2$, $x^{1/2 - \epsilon} = x^0 = 1$ and if $\epsilon > 1/2$, the exponent of $x$ is negative and the term $x^{1/2 - \epsilon}$ becomes a reciprocal of a fractional power of $x$. But when $\epsilon < 1/2$ it is not so clear if the terms converge to zero. [But see new edits added to the Derivation of Formula for $M(x)$ where L'Hospital's rule is applied to evaluate the limit.]
Derivation of Formula for M(x)
Mertens' function, $M(x)$, of a number $x$ is defined in terms of the sum of the values of $\mu(n)$ for all integers $\leq x$; however, the final value of $M(x)$ only depends on the odd square-free integers in the interval $(x/2, x]$ because for any integer $\leq x$ that contains a square factor $\mu(n) = 0$ and every even square-free integer $2n \leq x$ is derived from an odd square-free integer $n$ in the interval $[1, x/2]$; therefore, $\mu(n)$ has opposite sign to $\mu(2n)$ and they cancel. This fact implies that $M(x)$ equals the difference between the number of odd square-free integers in the interval $(x/2, x]$]that contain an even number of prime factors and those that contain an odd number of primes. So the value of $M(x)$ can be determined by counting the number of odd square-free integers $> x/2$ and $\leq x$ containing $1, 2, 3,... prime factors, multiplying each count by +/- 1 and adding these values.
The odd square-free integers $> x/2$ and $\leq x$ containing one prime factor are the primes, $p_i$, in this interval and $\mu(p_i) = -1$. The value these primes add to $M(x)$ equals $-[\pi(x) - \pi(\frac{x}{2})]$, where $\pi(x)$ is the prime counting function.
For an odd prime $p_i \leq \sqrt x$ the number of composite square-free integers $> x/2$ and $\leq x$ containing $p_i$ and a second prime is given by $[\pi(\frac{x}{p_i}) - \pi(\frac{x}{2p_i})$] if $x$ is sufficiently large (i.e., $\pi(\frac{x}{p_i})$ and $\pi(\frac{x}{2p_i})$ $\geq \pi(p_i)$).[This last condition is necessary to avoid including the square of $p_i$ in the count and double counting other integers containing $p_i$. If $x$ is finite these relations are modified] So the value that square-free integers of the form $p_{i}p_{j}$ add to the sum $M(x)$ is $+[\pi(\frac{x}{p_i}) - \pi(\frac{x}{2p_i})]$. A similar approach yields the values that odd square-free integers containing three or more prime factors ($p_i\cdots p_k$) add to the sum $M(x)$.
Combining all of the above steps gives the following relation to calculate the value of $M(x)$ if $x$ is sufficiently large (e.g., $x \to \infty$): \begin{equation} \begin{split} M(x) = -\left[\pi(x) - \pi\left(\frac{x}{2}\right)\right] + \left[\pi\left(\frac{x}{3}\right) - \pi\left(\frac{x}{2\cdot3}\right)\right] + \left[\pi\left(\frac{x}{5}\right) - \pi\left(\frac{x}{2\cdot5}\right)\right] + \hspace{2 in} \\ \cdots+ \left[\pi\left(\frac{x}{11}\right) - \pi\left(\frac{x}{2\cdot11}\right)\right] + \cdots - \left[\pi\left(\frac{x}{3\cdot5}\right) - \pi\left(\frac{x}{2\cdot3\cdot5}\right)\right] \cdots \hspace{2.2 in} \end{split} \end{equation}
Terms of the form $\pi(\frac{x}{p_i\cdots})$ approach $\frac{\pi(x)}{p_i\cdots}$ based on a result by Sierpinski: Let $a$ and $b$ be two positive real numbers, $0 < a < b$; it is easy to show that $\lim_{x \to \infty} \frac{log\ ax}{log\ bx} = 1$
Using this result and the prime number theorem, $\lim_{x \to \infty} \left(\pi(x)/\frac{x}{log(x)}\right) = 1$ one can derive the relation: $\pi(ax) \simeq \frac{a}{b} \pi(bx) \simeq a\pi(x) \ if \ b = 1$
Applying this result to equation (1) yields \begin{equation} \begin{split} M(x) \simeq -\left[\pi(x) - \frac{\pi(x)}{2}\right] + \left[\frac{\pi(x)}{3} - \frac{\pi(x)}{2\cdot3}\right] + \left[\frac{\pi(x)}{5} - \frac{\pi(x)}{2\cdot5}\right] \hspace{0.3 in}\\ +\left [\frac{\pi(x)}{7} - \frac{\pi(x)}{2\cdot7}\right] + \left[\frac{\pi(x)}{11} - \frac{\pi(x)}{2\cdot11}\right] +\cdots - \left[\frac{\pi(x)}{3\cdot5} - \frac{\pi(x)}{2\cdot3\cdot5}\right] \cdots \hspace{0 in} \end{split} \end{equation}
Removing the brackets from the terms in (2) and rearranging the terms in the natural order of the denominators gives the following expression: \begin{equation} \begin{split} M(x) \simeq -\pi(x) +\frac{\pi(x)}{2} + \frac{\pi(x)}{3} + \frac{\pi(x)}{5} - \frac{\pi(x)}{2\cdot3} + \frac{\pi(x)}{7} - \frac{\pi(x)}{2\cdot5} + \frac{\pi(x)}{11} \\ + \frac{\pi(x)}{13} - \frac{\pi(x)}{2\cdot7} - \frac{\pi(x)}{3\cdot5} \cdots - \frac{\pi(x)}{2\cdot11} \cdots + \frac{\pi(x)}{2\cdot3\cdot5} \cdots \hspace{1 in} \end{split} \end{equation}
Multiplying this relation by $-1$, factoring out $\pi(x)$ and taking the absolute value gives the expression: \begin{equation} |M(x)| \simeq \pi(x) \left(1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{5} + \frac{1}{2\cdot3} - \frac{1}{7} + \frac{1}{2\cdot5} - \frac{1}{11} - \frac{1}{13} + \frac{1}{2\cdot7} \cdots \right) \hspace{0 in} \end{equation} This last relation can be simplified using the Mobius function and the summation symbol to express the series in parentheses. \begin{equation} |M(x)| \simeq \pi(x)\sum_{n=1}^{\infty}\frac{\mu(n)}{n} \ as \ x \to \infty \hspace{2 in} \end{equation} Replacing $\pi(x)$ in (5) by ${x}/log(x)$ we obtain (6) \begin{equation} |M(x)| \simeq \frac{x}{log(x)}\sum_{n=1}^{\infty} \frac{\mu(n)}{n} \ as \ x \to \infty \hspace{1.8 in} \end{equation} Dividing both sides of (6) by $x^{1/2+\epsilon}$ gives the relation \begin{equation} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \frac{x^{1/2 - \epsilon}}{log(x)}\sum_{n=1}^{\infty} \frac{\mu(n)}{n} \hspace{2.4 in} \end{equation}
Von Mangoldt proved the series $\sum_{n=1}^\infty \mu(n)/n$ converges to zero and de la Vallee Poussin showed that for sufficiently large $x$, the series was less than $K/log(x)$. So replacing the summation by this factor gives \begin{equation} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \frac{x^{1/2 - \epsilon}}{log(x)}\frac{K}{log(x)} \ as \ x \to \infty \hspace{1 in} \end{equation}
In equation (8) if $\epsilon = 1/2$ the term $x^{1/2 - \epsilon} =x^0$ which equals 1 and the relation becomes \begin{equation} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \frac{1}{log(x)}\frac{K}{log(x)} \simeq 0 \ as \ x \to \infty \hspace{0.6 in} \end{equation} which approaches 0 since $1/log(x)$ and $K/log(x)$ approach 0 as $x \to \infty$.
If $\epsilon > 1/2$, the exponent $(1/2 - \epsilon)$ is negative so $x^{1/2 - \epsilon}$ equals a reciprocal of a power of $x$, call it $1/x^{\delta}$, where $\delta = \epsilon - 1/2$; so (8) can be written as \begin{equation} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \frac{1}{x^\delta}\frac{1}{log(x)}\frac{K}{log(x)} \simeq 0 \ as \ x \to \infty \hspace{0.5 in} \end{equation} which converges to 0 since $1/x^\delta$, $1/log(x)$ and $K/log(x)$ all converge to 0.
For the case $\epsilon < 1/2$ the limit of $\frac{|M(x)|}{x^{1/2 + \epsilon}} $ is of the form $\infty/\infty$ so the convergence can be evaluated by applying L'Hospital's rule twice and taking the derivative of the functions in the numerator and denominator as shown below: \begin{equation} \begin{split} \lim_{x \to \infty} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \lim_{x \to \infty} \frac{Kx^{1/2 - \epsilon}}{log^{2}(x)} \simeq \lim_{x \to \infty} \frac{K(1/2 - \epsilon)x^{-1/2 - \epsilon}}{2log(x)/x} \\ \simeq \lim_{x \to \infty} \frac{K(1/2 - \epsilon)x^{1/2 - \epsilon}}{2log(x)} \simeq \lim_{x \to \infty} \frac{K(1/2 - \epsilon)^{2}x^{-1/2 - \epsilon}}{2/x} \\ \simeq \lim_{x \to \infty} \frac{K(1/2 - \epsilon)^{2}x^{1/2 - \epsilon}}{2} \simeq \infty \hspace{1.5 in} \end{split} \end{equation} That the limit approaches $\infty$ is not what the author originally asserted and expected intuitively and leads to the startling--and unbelievable---conclusion that Riemann's hypothesis is not true.
Alternative Derivation of Formula for M(x)
This derivation is somewhat speculative in that I may be misusing the formula on which it is based, but I have attached it as evidence in support of the previous formula.
Let $\nu$ equal the number of distinct prime factors of an integer. The square-free integers consist of the primes ($\nu =1,\ \mu(n) = -1$), the composite integers containing an even number of prime factors ($\nu = 2, 4, 6,\dots,\ \mu(n) = +1$) and the composite integers containing an odd number of primes ($\nu = 3, 5, 7,\dots,\ \mu(n) = -1$).
Theoretically, $M(x)$ can be estimated by multiplying the number of square free integers with $\nu = 1, 3, 5,\dots$ by $-1$, then multiplying the number of square free integers with $\nu = 2, 4, 6,\dots$ by $+1$ and adding the two sets of numbers. (For convenience we omit $\mu(1) = +1$.) This approach is utilized below to obtain an asymptotic formula to estimate $M(x)$.
To estimate the number of square-free integers containing $\nu$ prime factors, $\pi_{\nu}(x)$, we use a formula proved by Sathe [Journal of the Indian Mathematical Society, Vol. 17(2), p. 63 (1953)]. For $\nu \sim kloglog(x)$ and $k = 1$, Sathe gives the following formula: \begin{equation} \pi_{\nu}(x) \sim \frac{6}{\pi^2}\frac{x}{lx}\frac{(l_2{x})^{\nu-1}}{(\nu-1)!} \hspace{1 in} \end{equation} where $lx = log(x)$, $l_{2}x = loglog(x)$ and $\pi = 3.14\cdots$.
Combining the above procedures gives the following relation as an estimate of $M(x)$: \begin{equation} \begin{split} M(x) \sim - \frac{6}{\pi^2}\frac{x}{lx}\frac{(l_2{x})^{1-1}}{(1-1)!} + \frac{6}{\pi^2}\frac{x}{lx}\frac{(l_2{x})^{2-1}}{(2-1)!} - \frac{6}{\pi^2}\frac{x}{lx}\frac{(l_2{x})^{3-1}}{(3-1)!} + \frac{6}{\pi^2}\frac{x}{lx}\frac{(l_2{x})^{4-1}}{(4-1)!} - \cdots \end{split} \end{equation} \begin{equation} \begin{split} \sim \frac{6}{\pi^2}\frac{x}{lx} \left[-1 + \frac{(l_2{x})^{1}}{(1)!} - \frac{(l_2{x})^{2}}{(2)!} + \frac{(l_2{x})^{3}}{(3)!} - \frac{(l_2{x})^{4}}{(4)!} + \cdots \right] \hspace{0.5 in} \end{split} \end{equation}
It is easy to show that the series within the brackets in this relation equals $-1$ times the Taylor series expansion $(e^{-u})$ where $u = l_{2}x$. So the expression in (4) can be written as \begin{equation} M(x)\sim -\frac{6}{\pi^2}\frac{x}{lx}[e^{-l_{2}x}] \sim -\frac{6}{\pi^2}\frac{x}{lx}\frac{1}{lx} \sim -\frac{6}{\pi^2}\frac{x}{(lx)^2} \ \ as\ x \to \infty \hspace{0 in} \end{equation} On taking the absolute value we get \begin{equation} |M(x)| \sim \frac{6}{\pi^2}\frac{x}{(lx)^2} \ \ as\ x \to \infty \hspace{1.3 in} \end{equation} Dividing both sides of this last equation by $x^{1/2 + \epsilon}$ (not shown) would give an expression comparable to the equation obtained previously: \begin{equation} \frac{|M(x)|}{x^{1/2 + \epsilon}} \simeq \frac{x^{1/2 - \epsilon}}{log(x)}\frac{K}{log(x)} \ as \ x \to \infty \hspace{1 in} \end{equation}