Show $\sum_{n\leq k\leq 2n}2^{-2k}\log(k)\leq C\, 2^{-2n}\log(n)$

Question

I'd like to prove

$$\sum_{n\leq k\leq 2n}2^{-2k}\log(k)\leq C \, 2^{-2n}\log(n),$$

where $C>0$ is a constant. Can someone give me a hint.

Answer 1

We can provide an integral form for such a sum by exploiting Frullani's theorem; partial summation is another way to provide tight bounds. Anyway, it is quite trivial that: $$\sum_{n\leq k\leq 2n}2^{-2k}\log k \leq \log(2n)\sum_{n\leq k\leq 2n}2^{-2k}\leq \log(2n)\sum_{k\geq n}2^{-2k} = \frac{4}{3}\log(2n) 2^{-2n} $$ that is $\leq \frac{8}{3}\log(n)2^{-2n}$ for any $n\geq 2$.