Proof for $p=1,2$
Euler's constant $\gamma=0.577215...$ is defined as [1]
\begin{equation}
\sum_{k=1}^{n} \frac{1}{k}\sim\ln n+\gamma ~~~~(1)
\end{equation}
Ramanujan claimed [2] that
\begin{equation}
\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0.~~~~(2)
\end{equation}
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])
\begin{equation}
\gamma=-\int_{0}^{\infty} \left( e^{-t}-\frac{1}{1+t} \right) \frac{dt}{t}.
\end{equation}
We can re-write it as
\begin{equation}
\gamma=-\lim_{x \to \infty} \left(\int_{1/x}^{x} \frac{e^{-t}-1}{t}+\int_{1/x}^{x} \frac{dt}{1+t}\right).
\end{equation}
\begin{equation}
\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),
\end{equation}
when $x$ is very large, we prove that
\begin{equation}
\sum_{k=1}^{\infty}(-1)^{k-1} \frac{x^k}{k (k!)} \sim \ln x+\gamma.
\end{equation}
Case 2: Let us use ([1]: p 955)
\begin{equation}
\gamma=1-\int_{0}^{\infty} \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}.
\end{equation}
\begin{equation}
\gamma=1-\lim_{x \to \infty} \int_{1/x}^{x} \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}
\end{equation}
\begin{equation}
\gamma= \lim_{x \to \infty} \left( 1+\ln x-\int_{1/x}^{x} \frac{\sin t}{t^2} dt\right)
\end{equation}
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 $-\nu-1 <\mu <1/2$.
\begin{equation}
\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)}~~~~(*).
\end{equation}
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
\begin{equation}
\gamma=\lim_{x \to \infty} \left( \ln 2x -\int_{1/x}^{x} \frac{J_0(t)}{t} dt\right).
\end{equation}
Finally, we use the series $J_0(x)=\sum_{0}^{\infty} (-1)^k \frac{(x/2)^{2k}}{(k!)^2}$, to write
\begin{equation}
\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).
\end{equation}
Lastly, we get
\begin{equation}
\sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x/2)^{2k}}{k (k!)^2} \sim 2 \ln (x/2)+2 \gamma,
\end{equation}
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.
