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Dec 19, 2021 at 5:04 answer added Claude Leibovici timeline score: 1
Jan 4, 2014 at 2:42 comment added Brendan McKay It's just a geometric series with the largest term at the end. Why don't you just sum it exactly? Then if you want an approximation for tiny $p$ use a Taylor expansion of the exact sum.
Jan 4, 2014 at 1:41 answer added Campello timeline score: 2
Jan 2, 2014 at 17:56 comment added Kelvin Lee My apologies, yes, I missed the exponent for $1-H(p)$, just edited my question.
Jan 2, 2014 at 17:55 history edited Kelvin Lee CC BY-SA 3.0
Fixed the typo discovered by the commenter.
Jan 2, 2014 at 8:03 comment added Andrew Is something missing in the sum? As wirtten it is $S(n)=\frac{n}{1-\operatorname{H}(p)}\,$, $f(n)=1-\operatorname{H}(p)$.
Jan 2, 2014 at 7:06 history asked Kelvin Lee CC BY-SA 3.0