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I need to compute $$f(x)=\sum_{i=0}^\infty \left(\left\lfloor\frac{i}{2^x}\right\rfloor+x+1\right)(1-p)^{i-1}p$$and minimize it with respect to $x$ (an expression which will depend on $p$).

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    You can pull the $1+x$ out of the sum. If $x \ne -1$ this will be infinite.2012-12-29
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    How could I possibly pull it out? $(1-p)^{i-1}*p$ is part of the sum2012-12-29
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    @Ross: You’re overlooking the effect of the $(1-p)^{i-1}$ factor. Presumably $0, so $\sum_i(x+1)(1-p)^{i-1}p$ is finite for all $x$.2012-12-29
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    yes. It would actually suffice to find a lower and an upper bound for the sum2012-12-29
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    @BrianM.Scott:I missed the $i$ in the exponent and thought that was outside the sum. Thanks.2012-12-29
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    Actually if I had $f(x)=x+1+\sum$, everything would be so much nicer.2012-12-29

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