# Series with exponential upper bound

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!

• i dont think so – 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.
• 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 $K \geq K_0 > 0$? – dstivd Jun 8 at 13:49