A dig at Ramanujan's: $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0$

Ramanujan's claim on page 98 in the book Ramanujan's note book part 1 by Bruce C. Berndt, is that $$\sum_{k=1}^{\infty} (−1)^{k−1}\frac{x^{pk}}{k(k!)^p}∼p\ln (x)+p\gamma,\quad p>0 \label{1}\tag{1}$$ The proposed proofs for $$p=1,2$$ are reported to be incorrect. For $$p>2$$ the claim \eqref{1} has been disproved. In the year $$1996$$, my proofs for $$p=1, 2$$ were evaluated to be correct by American Mathematical Monthly: however, similar proofs were told to have been published in the year 1995, somewhere. The result \eqref{1}, being of asymptotic kind, requires $$x$$ to be positive and large.

The question is here: what is the latest about this result when $$p\in (0,2]$$?

Any information or proof is welcome.

Proof for $$p=1,2$$
Euler's constant $$\gamma=0.577215...$$ is defined as [1] $$$$\sum_{k=1}^{n} \frac{1}{k}\sim\ln n+\gamma ~~~~(1)$$$$ Ramanujan claimed [2] that $$$$\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0.~~~~(2)$$$$ For $$p=1,2$$ the propose proofs were found incorrect. For $$p>2$$, the result (2) was disproved. Here the intent is to prove and show that for $$p=1,2$$ the result (2) is true and consistent with other existing results.
Case 1: Let us quote ([1]: p 955]) $$$$\gamma=-\int_{0}^{\infty} \left( e^{-t}-\frac{1}{1+t} \right) \frac{dt}{t}.$$$$ We can re-write it as $$$$\gamma=-\lim_{x \to \infty} \left(\int_{1/x}^{x} \frac{e^{-t}-1}{t}+\int_{1/x}^{x} \frac{dt}{1+t}\right).$$$$ $$$$\gamma=\lim_{x \to \infty}\left( \int_{1/x}^{x} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{t^{k-1}}{k!}dt-\ln x \right),$$$$ when $$x$$ is very large, we prove that $$$$\sum_{k=1}^{\infty}(-1)^{k-1} \frac{x^k}{k (k!)} \sim \ln x+\gamma.$$$$ Case 2: Let us use ([1]: p 955) $$$$\gamma=1-\int_{0}^{\infty} \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}.$$$$ $$$$\gamma=1-\lim_{x \to \infty} \int_{1/x}^{x} \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}$$$$ $$$$\gamma= \lim_{x \to \infty} \left( 1+\ln x-\int_{1/x}^{x} \frac{\sin t}{t^2} dt\right)$$$$ Introduce $$\sin x=2\sum_{n=0}^{\infty} (-1)^k J_{2k+1}(x)$$ ([1]: p 988]) and write $$\begin{eqnarray} \gamma= \lim_{x \to \infty} \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt\right)\\ \nonumber-2\sum_{k=1}^{\infty} \int_{0}^{\infty} (-1)^k \frac{J_{2k+1}(t)}{t^2}dt. \end{eqnarray}$$ Further by using ([1]: p 707), when $$-\nu-1 <\mu <1/2$$. $$$$\int_{0}^{\infty} x^\mu J_{\nu}(x) dx=2^{\mu} \frac{\Gamma(1/2+\nu/2+\mu/2)}{\Gamma(1/2+\nu/2-\mu/2)}~~~~(*).$$$$ we obtain $$\begin{eqnarray} \gamma{=}\lim_{x \to \infty} \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt\right)\\ \nonumber{-}\frac{1}{2} \sum_{k=1}^{\infty} (-1)^k \frac{1}{k(k+1)}. \end{eqnarray}$$ Let us integrate by parts taking $$J_1(x)$$ as first function and use $$2J_1'(x)=J_0(x)-J_1(x)$$. The infinite series occurring in above is nothing but $$1-2\ln 2$$. We get $$\begin{eqnarray} \gamma{=}\lim_{x \to \infty} \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt-\int_{1/x}^{x}\frac{J_1(t)}{t}dt\right)\\ \nonumber +\int_{0}^{\infty} \frac{J_1(t)}{t} dt-\frac{1}{2}+\ln 2. \end{eqnarray}$$ Note that $$J_1(x) \approx x/2$$, for small values of $$x$$ and for large values $$x \sim \infty, J_1(x)=\sqrt{\frac{2}{\pi x}} \cos (x-3\pi/4) \rightarrow 0$$. Using (*) again, we get $$$$\gamma=\lim_{x \to \infty} \left( \ln 2x -\int_{1/x}^{x} \frac{J_0(t)}{t} dt\right).$$$$ Finally, we use the series $$J_0(x)=\sum_{0}^{\infty} (-1)^k \frac{(x/2)^{2k}}{(k!)^2}$$, to write $$$$\gamma= \lim_{x \to \infty} \left ( \ln 2x -2 \ln x+\frac{1}{2} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x/2)^{2k}}{k(k!)^2} \right).$$$$ Lastly, we get $$$$\sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x/2)^{2k}}{k (k!)^2} \sim 2 \ln (x/2)+2 \gamma,$$$$ which is nothing but (2), for $$p=2$$.
$$[1]$$: I.S. Gradshteyn and I. M. Rhyzik, Table of Integrals", Series and Products (Academic Press(NY),1994).
$$[2]$$: Bruce C. Berndt, Ramanujan's Notebooks" Part 1, (Springer, Berlin, 1985) pp 98-100.