Let $K > 0$. Is it then true that there is some constant $C$ independent of $K$ such that $$\sum_{n=0}^\infty e^{-2^n K} \leq C e^{-K/C}$$

Thanks for the help!

  • $\begingroup$ i dont think so $\endgroup$ – mathworker21 Jun 8 at 12:11

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.

  • $\begingroup$ Thanks for the answer! That makes sense. Do you think the inequality would hold if I bound K away from 0. So let's say $K \geq K_0 > 0$? $\endgroup$ – dstivd Jun 8 at 13:49
  • $\begingroup$ Not sure. I'll have to think about it. $\endgroup$ – saulspatz Jun 8 at 14:10
  • $\begingroup$ Actually, I think according to the site rules, you should ask this as another question. It's probably a good idea anyway, to attract more attention. $\endgroup$ – saulspatz Jun 8 at 14:12

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