Problem source

Show that the sum of the infinite series 
\[
\log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots
\]
is 
\[
\frac{1}{\ln(2\sqrt{2})}.
\]
{[}$\log_{a}b=c$ is equivalent to $a^{c}=b$.{]} 

Solution source

Let $S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots$ then
\begin{align*}
S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\
&= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\
&= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\
&= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\
&= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\
&= \frac{1}{\ln (2^{3/2})} \\
&= \frac{1}{\ln (2 \sqrt{2})}
\end{align*}