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Double series

Evaluate the series:

\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}

(Putnam Competition , 2016)

Solution

Since the double series the series converges absolutely we can interchange the summation. Thus:

\begin{aligned} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \frac{1}{k2^n+1} &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \int_{0}^{1} x^{k 2^n } \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^{2^n k} \, {\rm d}x\\ &= \sum_{n=0}^{\infty} \int_{0}^{1} \log \left ( 1+x^{2^n} \right ) \, {\rm d}x\\ &= \int_{0}^{1}\sum_{n=0}^{\infty} \log \left ( 1+ x^{2^n} \right ) \, {\rm d}x \\ &= \int_{0}^{1} \log \prod_{n=0}^{\infty} \left ( 1+x^{2^n} \right ) \, {\rm d}x \\ &= - \int_{0}^{1} \log \left ( 1-x \right ) \, {\rm d}x \\ &=1 \end{aligned}

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