Math 01

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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$$\log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots = \sum_{k=1}^{\infty} (-1)^{k}\log_{2^{2^{k}}}\mathrm{e}$$

$$= \sum_{k=1}^{\infty} \frac{(-1)^{k}}{{2^{k}}\ln{2}}$$

$$= \frac{1}{\ln{2}}\sum_{k=1}^{\infty} \frac{(-1)^{k}}{2^{k}} = \frac{1}{\ln{2}}\sum_{k=1}^{\infty} {(-\frac{1}{2})}^{k}$$

$$= \frac{1}{\ln{2}} \cdot \frac{1}{1 - (-\cfrac{1}{2})} = \frac{1}{\cfrac{3}{2} \cdot \ln{2}}$$

$$= \frac{1}{\ln(2\sqrt{2})}$$