I didn't make any headway on this during the Putnam, mainly because I wasn't going to waste my time on a B6 question which should be the most difficult one on the exam. It seems really intriguing, and was wondering if anyone had any good ideas on how to tackle it or a solution:$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{1}{k2^n+1}=$$
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1$\begingroup$ What is the question? $\endgroup$– BernardDec 4, 2016 at 2:11
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$\begingroup$ I'm assuming what it converges to, but it literally was just that and no words $\endgroup$– Drunk DerivingDec 4, 2016 at 2:12
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2$\begingroup$ I put good odds on getting somewhere by replacing $2^n$ with $x^n$ and doing some calculus. $\endgroup$– user14972Dec 4, 2016 at 2:56
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1$\begingroup$ I answered by factoring a $k 2^n$ out of the denominator of the second summand to make it into a convergent geometric series, then after playing around with sums and combining things to simplify, it ends up as a telescoping sum that evaluates to 1. $\endgroup$– Paul LeVanDec 4, 2016 at 6:42
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4$\begingroup$ I would let $a(k)$ represent the inner series, and show that the above series is the same as $\sum\limits_{k = 1}^\infty \frac{a(k) - a(2k)}{k}$. The expression $a(k) - a(2k)$ gives a telescoping series with sum $1/(k+1)$, so the result is $\sum\limits_{k = 1}^\infty \frac{1}{k(k+1)} = 1$. $\endgroup$– kobeDec 4, 2016 at 17:10
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1 Answer
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HINT: Let $S_{0}$ and $S_{1}$ be the sums $\sum_{k}\frac{1}{k}\sum_{n=0}^{\infty} \frac{1}{k\times2^n+1}$ with $k$ running over all odd and even positive integers respectively so that we have, $$S = S_1-S_0 = (S_1+S_0)-2S_0...(1)$$ Thus, we can write $k =2\alpha$ for $S_1$ to get $$S_1 = \sum_{\alpha=1}^{\infty}\frac{1}{2\alpha} \sum_{n=0}^{\infty}\frac{1}{\alpha2^{n+1}+1}$$ $$=\frac{1}{2}(S_0+S_1) + \sum_{\alpha=1}^{\infty}\frac{1}{2\alpha(\alpha+1)}$$ $$=\frac{1}{2}(S_0+S_1) + \frac{1}{2}$$ as the last sum telescopes which can checked very easily giving us $1/2$. This gives us $S=1$ using $(1)$. Hope it helps.