Question :
My attempt:
Let $a=17399172$ $$\begin{align} &\sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(\ln\left(e^{C_j^k}\right)\right)\right)(1-e)^k} \\ &= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(C_j^k\right)\right)(1-e)^k} \\ &= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(2^k\right)(1-e)^k} \\ &= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k} \\ \end{align}$$
Can you give me some hints what to do afterwards? I'll try by myself.
This is not a full answer but I will elaborate on @Dr.WolfgangHintze comments. We have $$ \begin{align} S &= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}\\ &= -\sum_{k=0}^{a-1}(1+H_{k+1})\frac{(\frac{1}{1-e})^{k+1}}{k+1} \\ &= -\int_0^{\frac{1}{1-e}}\sum_{k=0}^{a-1}(1+H_{k+1})t^k\,\mathrm dt\\ &=: -\int_0^{\frac{1}{1-e}}S^\ast(t)\,\mathrm dt. \end{align} $$ Now write $$ \begin{align} S^\ast(t) &= \sum_{k=0}^{a-1}(1+H_{k+1})t^k\\ &= \frac{1-t^a}{1-t}+\sum_{k=0}^{a-1}H_{k+1}t^k\\ &= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=a}^\infty H_{k+1}t^k\\ &= \frac{1-t^a}{1-t}+\sum_{k=0}^\infty H_{k+1}t^k-\sum_{k=0}^\infty H_{a+k+1}t^{a+k}. \end{align} $$ Working with the recurrence relations for the harmonic numbers as well as the Cauchy product we find $$ \begin{align} \sum_{k=0}^\infty H_{a+k+1}t^{a+k} &=t^a\sum_{k=0}^\infty (H_a+\sum_{\ell=1}^{k+1}\frac{1}{a+\ell})t^k\\ &=t^a\left(\frac{H_a}{1-t}+\sum_{k=0}^\infty\sum_{\ell=0}^k\frac{1}{a+1+\ell}t^k\right)\\ &=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{1}{a+1+k}t^k\right)\\ &=\frac{t^a}{1-t}\left(H_a+\sum_{k=0}^\infty\frac{(1)_k\Gamma(a+1+k)}{\Gamma(a+2+k)}\frac{t^k}{k!}\right)\\ &=\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right). \end{align} $$ Setting $a=0$ gives the closed form for the other series term and so $$ \begin{align} S^\ast(t) &= \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right). \end{align} $$ Therefore, $$ \begin{align} S &= -\int_0^{\frac{1}{1-e}} \frac{1-t^a}{1-t}-\frac{\log(1-t)}{1-t}-\frac{t^a}{1-t}\left(H_a+\frac{1}{1+a}F\left({1,1+a\atop 2+a};t\right)\right)\,\mathrm dt. \end{align} $$ Most of the integrals are easy to evaluate with the exception of the integral involving the hypergeometric function which will take a little extra work.