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 $a^{c}=b$ ]
STEP I 1987 Question 4 (Pure)
Solution
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$$ \log{2}\mathrm{e}-\log{4}\mathrm{e}+\log{16}\mathrm{e}-\ldots+(-1)^{n}\log{2^{2^{n}}}\mathrm{e}+\ldots = \sum{k=1}^{\infty} (-1)^{k}\log{2^{2^{k}}}\mathrm{e} $$
$$ = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{{2^{k}}\ln{2}} $$
$$ = \frac{1}{\ln{2}}\sum{k=1}^{\infty} \frac{(-1)^{k}}{2^{k}} = \frac{1}{\ln{2}}\sum{k=1}^{\infty} {(-\frac{1}{2})}^{k} $$
$$ = \frac{1}{\ln{2}} \cdot \frac{1}{1 - (-\cfrac{1}{2})} = \frac{1}{\cfrac{3}{2} \cdot \ln{2}} $$
$$ = \frac{1}{\ln(2\sqrt{2})} $$