It's not e

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {n-1}{n!} \\ & = \sum_{\blue{n=1}}^\infty \frac n{n!} - \sum_{\blue{n=1}}^\infty \frac 1{n!} \\ & = \frac d{dx} \sum_{\red{n=0}}^\infty \frac {x^n}{n!} \bigg|_{x=1} - \sum_{\red{n=0}}^\infty \frac 1{n!} \red{+1} \\ & = \frac d{dx} e^x \bigg|_{x=1} - e + 1 \\ & = e-e+1 \\ & = 1 \end{aligned}$

Therefore $\log_\frac 12 (4S) = \dfrac {\log 4}{-\log 2} = -\dfrac {2\log 2}{\log 2} = \boxed {-2}$ .

x/1!+x^2/2!+x^3/3!+...=e^x-1 then dividing both sides by x yields, x^0/1!+x/2!+...=(e^x-1)/x.Now taking derivative respect to x gives, (1-1)x^-1/1!+1x^0/2!+2x/3!+... =(1-e^x)/x^2+e^x/x.Setting x=1, S=1.So log(base 1/2)(4)=-2.

We can break down S as follows:

S = $\sum_{ i=1 }^{∞}$ $\frac{n-1}{n!}$

= $\sum_{ i=1 }^{∞}$ $\frac{n}{n!}$ - $\sum_{ i=1 }^{∞}$ $\frac{1}{n!}$

= $\sum_{ i=1 }^{∞}$ $\frac{1}{(n-1)!}$ - $\sum_{ i=1 }^{∞}$ $\frac{1}{n!}$

= $\sum_{ i=0 }^{∞}$ $\frac{1}{n!}$ - $\sum_{ i=1 }^{∞}$ $\frac{1}{n!}$

= $\frac{1}{0!}$ = 1

As S = 1, We have $\ Log_{1/2}{(4×1)}$

= $\ Log_{1/2} {1/2} ^{-2}$

= -2