Is the alternating sum of prime numbers $2-3 + 5-7 \dots$ asymptotic to $\frac{1}{2}p_k$?

I tried to show that $|2-3 + \dots (-1) ^ {N + 1} p_N| \sim \frac{1}{2}p_N$ but I don't know if what I did holds. I would be very interested in getting advice. At the end of the question there is a bonus speculation which is both very stupid and interesting, at least, from my point of view...

$p_k$ is the $k$-th prime number, expanding the inverse function of the logarithmic integral for $k\geq2$ we get: $$p_k= k \ln(k)+k \ln(\ln(k))-k+O\left(\frac{k }{\ln(k)}\ln(\ln(k))\right)$$

I wonder if it is correct to sum over the big $O$ notation and exploit the algebra of this notation $$O\left((-1)^{N+1}\frac{N \ln\ln N}{\ln(N)}\right)+O\left((-1)^{N}\frac{(N-1) \ln\ln (N-1)}{\ln(N-1)}\right)+\dots=O\left(\frac{(N) \ln\ln (N)}{\ln(N)}\right)$$

If this is legit, we can write $$\sum_{k=1}^N (-1)^{k+1}p_k=\sum_{k=1}^N (-1)^{k+1} k \ln(k)+\sum_{k=1}^N (-1)^{k+1} k \ln\ln(k)-\sum_{k=1}^N (-1)^{k+1} k +O\left(\frac{N \ln\ln N}{\ln(N)}\right)$$

To estimate the value of these sums, let's set $ N = 2x + 1 $ $$\sum_{k=1}^{2x+1}(-1)^{k+1}k \ln(k)=\sum_{k=1}^{x+1} (2k-1) \ln(2k-1)-\sum_{k=1}^x 2k \ln(2k)=\\\sum_{k=1}^{x+1} 2k \ln(1-\frac{1}{2k})-\sum_{k=1}^{x+1} \ln(2k-1)+2(x+1)\ln(2(x+1)).$$

$$\sum_{k=1}^{x+1} \ln(2k-1)=(x+1)\ln(2)+\ln\left((x+\frac{1}{2})!\right)-\frac{1}{2}\ln(\pi)$$

The Lauren series for $y \ln(1-\frac{1}{y})$ for $y\to \infty$ is:$\ \ -1-\frac{1}{2y}-\frac{1}{3y^2}-\frac{1}{4y^3}+O(y^{-4})$

$$\sum_{k=1}^{x+1} 2k \ln(1-\frac{1}{2k})\sim -(x+1)-\frac{1}{4}H_{x+1}-\sum_{k=2}^{\infty}\frac{\zeta(k)}{2^k(k+1)}$$ $H$ is the Harmonic number $H_{x+1}\sim\ln(x+1)+\gamma$ $$\sum_{k=1}^{x+1} 2k \ln(1-\frac{1}{2k})=-x+O(\ln(x))$$

By the Stirling's approximation : $\ln(y!)\sim y \ln(y)-y+O(\ln(y))$

$$\sum_{k=1}^{x+1} \ln(2k-1)= x (\ln(x)+\ln(2)-1)+O(\ln(x))$$

$$\left|\sum_{k=1}^{2n+1}(-1)^{k+1} k \ln(k)\right|\sim n \ln(2n)+O(\ln(n))$$ In the same way we can estimate $$\left|\sum_{k=1}^{2n+1}(-1)^{k+1}k \ln(\ln(k))\right|= n\ln(\ln(2n))+O\left(\ln(\ln(n))\right)$$ and $$\left| \sum_{k=1}^{2n+1} (-1)^{k+1}k\right|=n+1$$

$$ \left|\sum_{k=1}^{2n+1}(-1)^{k+1}p_k\right|=n\Big( \ln(2n)+\ln(\ln(2n)-1\Big)+O\left(\frac{n \ln\ln(n)}{\ln(n)}\right) $$

or in other words

$$ \left|\sum_{k=1}^{N}(-1)^{k+1}p_k\right|=\frac{N}{2}\Big( \ln(N)+\ln(\ln(N)-1\Big)+O\left(\frac{N \ln\ln(N)}{\ln(N)}\right) .$$

Basically this means that the absolute value of the sum oscillates around $\frac{P_N}{2}$

$$S(x)=\left|\sum_{k=1}^{x}(-1)^{k+1}p_k\right|-\frac{p_x}{2}$$ Therefore the function $S(x)$ will have no limit for $x \to\infty$ but will oscillate around the abscissa axis as if it were a sound wave. Who knows, maybe there is some way to code the $S$ function in some music software to know how the prime numbers sounds like.

Mathematicians believe that It should by asymptotic to $\frac{p_x}{2}$ but its not proven, the best unconditional lower bound is $\frac{p_x}{64}$ and the conditional lower bound is $\frac{p_x}{8}$ assuming Hardy-Littlewood conjecture.

The problem with your analysis is that if $f(x) \sim g(x)$ that does not mean that $ |\sum \limits_{k=1}^{n} (-1)^{k} f(k) | \sim |\sum \limits_{k=1}^{n} (-1)^{k} g(k) |$ for instance $f(x)=x+(-1)^{\lfloor x \rfloor} \sqrt{x}$ and $g(x)=x$

And so $ |\sum \limits_{k=1}^{n} (-1)^k g(k)| \sim \frac{n}{2} \not \sim \frac{n \sqrt{n}}{6} \sim |\sum \limits_{k=1}^{n} (-1)^k f(k)|$

The summation in this question boils down to the gap between consecutive primes $g_k$ which we don't know that much about it.