Let $K > 0$. Is it then true that there is some constant $C$ independent of $K$ such that $$\sum_{n=0}^\infty e^{-2^n K} \leq C e^{-K/C}$$
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2026-03-25 05:03:58.1774415038
Series with exponential upper bound
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No. If it were true, we would have $$\sum_{n=0}^\infty e^{-2^n K} < C,\ K>0$$because $e^{-K/C}<1.$ This is not true, since by letting $K\to0+,$ we can make the sum of the first $n$ terms of the left-hand side approach $n$ as nearly as desired.