# Putnam_2016_B6 this was v nice xD

we have S=\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}{\sum_{n=0}^{\infty}\frac{1}{k2^n+1}}.
which is obviously reminiscent of the maclaurin series expansion of \displaystyle \ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}x^k for x\in(-1,1]
comparing the above 2 and noting their similarity, we want somehow to express \displaystyle\sum_{n=0}^{\infty}\frac{1}{k2^n+1} in terms of (A(x))^k for some function A(x).
consider \displaystyle\int x^{k2^n}dx=\frac{x^{k2^n+1}}{k2^n+1}\implies\sum_{n=0}^{\infty}\frac{1}{k2^n+1}=\sum_{n=0}^{\infty}\int_{0}^{1} x^{k2^n}dx
so this means we now have (the integral of) (A(x))^k=x^{k2^n} in the expression for S, as we now know \displaystyle S=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\int_{0}^{1} x^{k2^n}dx
\displaystyle\int_{0}^{1} x^{k2^n}dx=\frac{1}{k2^n+1}<\frac{1}{k2^n}<\frac{1}{2^n} \ \forall \ n\in\mathbb{Z}_0^+, k\in\mathbb{Z}^+
\implies\displaystyle\sum_{n=0}^{\infty}\int_{0}^{1} x^{k2^n}dx<\sum_{n=0}^{\infty}\frac{1}{2^n}=2 \implies\sum_{n=0}^{\infty}\int_{0}^{1} x^{k2^n}dx must also converge.

\begin{align} \implies S&=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\int_{0}^{1} \frac{(-1)^{k-1}}{k}(x^{2^n})^kdx \\ &= \sum_{n=0}^{\infty}\int_{0}^{1} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}(x^{2^n})^kdx \end{align}

interchanging order of k summation and integral summation with n, ie. \displaystyle\sum_{k=1}^{\infty} with \displaystyle\sum_{n=0}^{\infty}\int_{0}^{1}, is valid due to convergence of integral summation with n as proven.
x only ranges between limits of integral, so x\in[0,1]\implies x^{2^n}\in[0,1].
\displaystyle\implies S=\sum_{n=0}^{\infty}\int_{0}^{1} \ln(1+x^{2^n})dx
considering integral as area under graph of \ln(1+x^{2^n}) between 0 and 1, the curve will be continuous and increasing meaning the area is finite, and as n increases the curve will flatten and so the area will decrease (crucially remaining finite)
(you can verify this on desmos at https://www.desmos.com/calculator/2eyyetuiwq)
so integral is convergent, and we can interchange order of integration and summation.

\begin{align} \implies S&=\int_{0}^{1}\sum_{n=0}^{\infty} \ln(1+x^{2^n})dx \\ &=\int_{0}^{1}\ln(\prod_{n=0}^{\infty}(1+x^{2^n}))dx \end{align}

notice that

\begin{align} \prod_{n=0}^{\infty}(1+x^{2^n})&=(1+x)(1+x^2)(1+x^4)(1+x^8)\ldots \\ &=1+x+x^2+x^3+x^4+\ldots \\ &=\sum_{n=0}^{\infty}x^n \\ &=(1-x)^{-1} \end{align}
\begin{align} \therefore S&=\int_{0}^{1}\ln((1-x)^{-1})dx \\ &=\int_{1}^{0}\ln(1-x)dx \\ &=\left[(x-1)ln(1-x)\right]_1^0 - \int_1^0 \frac{-(x-1)}{1-x}dx \\ &=(0-0) - \left[x\right]_1^0 \\ &=-(0-1) \\ &=1 \end{align}
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