Upper bound of $\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(1+x)(1+2x)(1+3x)\cdots(1+nx)} $ $$
\begin{aligned}
\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(1+x)(1+2x)(1+3x)\cdots(1+nx)} && x > 0
\end{aligned}
$$
I tried using (1) some inequalities (2) Taking the coefficients of x common to get some factorials in the denominator , couldn't reach a right conclusion. One more school of thought: as the series is convergent so its upper bound is nothing but the sum itself. No idea what I am missing! Any help?

Where the problem is exactly happening in the expression in the above image, although I know I am way far in reaching a right conclusion.
 A: Given the commented multiple-choice answers, another approach would be to notice the following:
$$\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(1+x)(1+2x)(1+3x)\cdots(1+nx)}\\
\le \sum_{n=0}^{\infty}\frac{x^{2n}}{(1+x)(1+2x)(1+3x)\cdots(1+nx)}\\
\le \sum_{n=0}^{\infty}\frac{x^{2n}}{(x)(2x)(3x)\cdots(nx)}\\
=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=e^x$$
A: We can try to get a closed form for the sum, what will simplify the further analysis.
$$S(x)=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(1+x)(1+2x)\cdots(1+nx)}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{n}}{(1/x+1)(1/x+2)\cdots(1/x+n)}$$
$$=\sum_{n=0}^{\infty}\frac{(-1)^n x^{n}\,\Gamma(1/x+1)}{\Gamma(1/x+1)(1/x+1)(1/x+2)\cdots(1/x+n)}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{n}\,\Gamma(1/x+1)}{\Gamma(1/x+1+n)}$$
$$=\sum_{n=0}^{\infty}\frac{(-1)^n x^{n-1}\,\Gamma(1/x)}{\Gamma(1/x+1+n)}\frac{\Gamma(n+1)}{\Gamma(n+1)}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{n-1}\,\Gamma(1/x)\Gamma(n+1)}{n!\,\Gamma(1/x+1+n)}$$
$$\frac{1}{x}\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}B(1/x;n+1)$$
Using the integral representation of the beta-function
$$S(x)=\frac{1}{x}\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}\int_0^1t^{1/x-1}(1-t)^ndt$$
Changing the order of summation and integration
$$S(x)=\frac{1}{x}\int_0^1t^{1/x-1}dt\sum_{n=0}^{\infty}\frac{(-1)^n x^n\,(1-t)^n}{n!}=\frac{1}{x}\int_0^1t^{1/x-1}e^{-x(1-t)}dt$$
$$\boxed{\,\,S(x)=\frac{e^{-x}}{x}\int_0^1t^{1/x-1}e^{xt}dt\,\,}\qquad(1)$$
Now we can consider different limits.
At $x\to0$
$$S(x)\to\frac{e^{-x}}{x}\int_0^1t^{1/x-1}\Big(1+O(tx)\Big)dt=e^{-x}\Big(1+O(x)\Big)\to 1$$
At $x\to\infty$ , integrating (1) by part
$$S(x)=e^{-x}t^{1/x}e^{xt}\Big|_{t=0}^{t=1}-e^{-x}x\int_0^1t^{1/x}e^{xt}dt=1-x\int_0^1(1-s)^{1/x}e^{-xs}ds$$
$$=1-\int_0^x\big(1-t/x\big)^{1/x}e^{-t}dt=1-\int_0^xe^{-t}e^{\frac{1}{x}\ln\big(1-\frac{t}{x}\big)}dt$$
$$=1-1+\frac{1}{x^2}\int_0^\infty te^{-t}dt+O(1/x^3)=\frac{1}{x^2}+O(1/x^3)\to 0$$
