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Let Hn=1+12+13+⋯+1n H_n = 1+\frac12 + \frac13 + \cdots + \frac1{n}Hn=1+21+31+⋯+n1 be the nthn^\text{th}nth harmonic number.
The sum ∑n=1∞Hn2n=ln(a)\displaystyle \sum_{n=1}^\infty \frac{H_n}{2^n} = \ln (a) n=1∑∞2nHn=ln(a) for some positive integer a. a.a.
What is a? a?a?
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