Is That Too Much to Calculate? (3)

First note that $1 - \sum_{n=0}^{k-1}\frac{e^{-t}t^n}{n!} = e^{-t}\left(e^t - \sum_{n=0}^{k-1} \frac{t^n}{n!}\right) = e^{-t}\sum_{n=k}^\infty \frac{t^n}{n!}$ so that $\begin{aligned} \sum_{k=1}^\infty \left(1-\sum_{n=0}^{k-1}\frac{e^{-t}t^n}{n!}\right)\big(1-p\big)^{k-1}p &= e^{-t}\sum_{k=1}^\infty \left(\sum_{n=k}^\infty \frac{t^n}{n!}\right)\big(1-p\big)^{k-1}p \\ &= e^{-t}\sum_{1 \leq k \leq n} \left(\frac{t^n}{n!} \cdot (1-p)^{k-1}p\right) \\ &= e^{-t}\sum_{n=1}^\infty \frac{t^n}{n!} \sum_{k=1}^n (1-p)^{k-1}p \\ &= e^{-t}\sum_{n=1}^\infty \frac{t^n}{n!} \cdot \big(1 - (1-p)^n\big) \\ &= e^{-t}\left(\left(\sum_{n=1}^\infty \frac{t^n}{n!}\right) - \left(\sum_{n=1}^\infty \frac{\big(t(1-p)\big)^n}{n!}\right)\right) \\ &= e^{-t}\left(\left(e^t - 1\right) - \left(e^{t(1-p)} - 1\right)\right) \\ &= 1 - e^{-tp} \end{aligned}$

Setting $t=2018$ and $p=\ln 2$ yields the answer of $1 - e^{-2018\ln 2} = 1 - \left(\frac{1}{2}\right)^{2018}$

$\begin{aligned} S & = \sum_{k=1}^\infty \left(1-\sum_{n=0}^{k-1}\frac {e^{-t}t^n}{n!}\right) (1-p)^{k-1}p \\ & = p\sum_{\color{#3D99F6}k=1}^\infty (1-p)^{\color{#3D99F6}k-1} - pe^{-t} \sum_{k=1}^\infty (1-p)^{k-1} \sum_{n=0}^{k-1}\frac {t^n}{n!} \\ & = p\sum_{\color{#D61F06}k=0}^\infty (1-p)^{\color{#D61F06}k} - pe^{-t} \left(1+(1-p)(1+t) + (1-p)^2 \left(1+t+\frac {t^2}{2!} \right) + (1-p)^3 \left(1+t+\frac {t^2}{2!} + \frac {t^3}{3!} \right) + \cdots \right) \\ & = \frac p{1-(1-p)} - pe^{-t} \left(\frac 1{0!} \sum_{n=0}^\infty (1-p)^n + \frac t{1!} \sum_{n=1}^\infty (1-p)^n + \frac {t^2}{2!} \sum_{n=2}^\infty (1-p)^n + \frac {t^3}{3!} \sum_{n=3}^\infty (1-p)^n + \cdots \right) \\ & = 1 - pe^{-t} \left(\frac 1{0!p} + \frac {(1-p)t}{1!p} + \frac {(1-p)^2t^2}{2!p} + \frac {(1-p)^3t^3}{3!p} + \cdots \right) \\ & = 1 - e^{-t} e^{(1-p)t} = 1 - e^{-t+t-pt} = 1 - e^{-pt} = 1 - \frac 1{e^{2018\ln 2}} = \boxed{1-\dfrac 1{2^{2018}}} \end{aligned}$