Funky Series 2

$S=\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k~(\log_{2017} 2018)^k}{k!~(\log_{2017} 2016)^k}$

$\quad =\displaystyle\sum_{k=0}^{\infty}\left(\frac{-1\cdot\log_{2017} 2018}{\log_{2017} 2016}\right)^k\cdot\frac{1}{k!}$

$\quad =\displaystyle\sum_{k=0}^{\infty}\color{#D61F06}{\left(\frac{\log_{2017}(\frac{1}{2018})}{\log_{2017} 2016}\right)}^k$ $\cdot\dfrac{1}{k!}$

Using the reverse change-of-base formula, we can rewrite what's inside the parentheses:

$\quad =\displaystyle\sum_{k=0}^{\infty} \color{#D61F06}{\left(\log_{2016} \left(\frac{1}{2018}\right)\right)^k}$ $\cdot\dfrac{1}{k!}$

If we plug in $x=\log_{2016}(\frac{1}{2018})$ into the series, we have the series of the form $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$ for all real $x$ .

Therefore, $S=e^x=e^{\log_{2016}(\frac{1}{2018})} \implies \ln S =\log_{2016}(\frac{1}{2018})=-\log_{2016}(2018)$ .

Thus, $a=2016, b = 2018 \implies \frac{a+b}{2} = \boxed{2017}$ .

$\begin{aligned} S & = \sum_{k=0}^\infty \frac {(-1)^k (\log_{2017} 2018)^k}{k! (\log_{2017} 2016)^k} & \small \color{#3D99F6} \text{Let }x = \frac {\log_{2017} 2018}{\log_{2017} 2016} \\ & = \sum_{k=0}^\infty \frac {(-1)^k x^k}{k!} \\ & = e^{-x} \\ & = \exp \left(- \frac {\log_{2017} 2018}{\log_{2017} 2016} \right) \\ & = \exp \left(- \frac {\frac {\ln 2018}{\ln 2017}}{\frac {\ln 2016}{\ln 2017}} \right) \\ & = \exp \left(- \frac {\ln 2018}{\ln 2016} \right) \end{aligned}$

$\begin{aligned} \implies \ln S & = - \frac {\ln 2018}{\ln 2016} = - \log_{2016} 2018 \end{aligned}$

Therefore, the average of $a$ and $b$ is $\dfrac {a+b}2 = \dfrac {2016+2018}2 = \boxed{2017}$