The prime number theorem in its usual form is somewhat obvious and maybe intuitive from tables of data that prompted Gauss in 1792 or 1793 to speculate that the density of primes was $1/\log x.$ Unfortunately very little is obvious beyond this. Proofs of the PNT often use the version $\psi(x)\sim x$ and so intuition about proofs of the PNT require some familiarity with that version, which can be shown to be equivalent to the usual $\pi(x)\sim x/\log x.$ See Apostol [1], p. 79.
Your question (in bold) is, what does the implication $\sum \mu(x)/x=0\leftrightarrow PNT$ look like? I mentioned in a comment that it depends on the result that $\sum \mu(x)=o(x) \leftrightarrow PNT.$ Because this latter result is easy to find (see, for example, Apostol), below is only a sketch of the last step from Ayoub [1], showing that $\sum \mu(x)/x= o(1) \leftrightarrow \sum \mu(x)=o(x)$ (and hence $\sum \mu(x)/x=o(1)\leftrightarrow PNT$).
The general theorem below is a prerequisite for the proof and I think Apostol and others more or less also prove it so the proof is omitted.
Theorem. If $a(x)$ is defined for integral $x$, if B(x) is of bounded variation in every finite interval, if $\sum_{n \leq x}a(n)=o(x), if B(x) =O(1)$ and $\sum_{n\leq x}|a(n)|=O(x)$ then
$$\sum_{n\leq x}a(n)B\left(\frac{x}{n}\right)=o(x) $$ [proof omitted].
Main result. Let $M(x) = \sum_{n\leq x}\mu(n), L(x) =\sum_{n\leq x}\frac{\mu(n)}{n}.$ Then $M(x) =o(x)$ if and only if $L(x)=o(1).$
First assume $M(x)=o(x).$ In the theorem above put $a(n)=\mu(x)$ and $B(x)=x-[x].$ Then
$$\sum_{n\leq x }\mu(n)\left( \frac{x}{n}-\left[\frac{x}{n}\right]\right)=o(x). $$
But $\sum_{n\leq x}\mu(n)\left[ \frac{x}{n}\right]=1$ (see Apostol, etc.) and so
$$\sum_{n\leq x}\mu(n)\frac{x}{n}=o(x)$$ and dividing by $x,~L(x)=o(1).$
Now assume $L(x) = o(1).$ Then $M(x)=\sum_{n\leq x}\mu(n)=\sum \frac{\mu(n)\cdot n}{n}$
$$= xL(x)-\int_1^x L(t)~ dt = o(x) + o(x) =o(x). $$
The last line is a consequence of the following lemma.
Lemma. Let $x\geq 1$ and $\phi(x)$ have continuous derivatives for $x\geq 1.$ Let $S(x)=\sum_{n\leq x}C(n)$ with $C(n)$ real or complex numbers. Then
$$\sum_{n\leq x}C(n)\phi(n)=S(x)\phi(x)-\int_1^x S(t)\phi'(t)dt. $$
If we put $C(n)=\mu(n)/n$ and $\phi(x)=x$ then the last line next above follows. This is the direction of the proof that is included in Apostol's text [2], so you can get the non-trivial details there on page 97.
[1] Ayoub, An Introduction to the Theory of Numbers, AMS 1963.
[2] Apostol, Introduction to Analytic Number Theory, Springer 2000.
Best Answer
Let $n$ be an odd integer. Then, there exists an integer $k$ such that $n = 2k + 1$. Thus, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$. Since there exists an integer $l$ such that $n^2 = 2l + 1$, namely, $l = 2k^2 + 2k$, $n^2$ is odd.