Summation of logarithmic series

Question

I am solving a recurrence relation and it requires me to sum the following series upto $\log{n}$ terms -
$1/\log(n) + 1/\log(n/2) + 1/\log(n/4)$.....

The base in each term is $2$. Any help on how to proceed?

Answer 1

For every $n$, one considers considers $$s(n)=\sum_{k\geqslant0}\frac1{\log_2(n/2^k)}, $$ where the sum is over every $k$ such that $2^k\lt n$ and $\log_2$ is the logarithm with base $2$. One cannot hope any explicit formula valid for every $n$ but note that, for every $i$, $$s(2^i)=\sum_{k=0}^{i-1}1/(i-k)=H_i,$$ where $H_i$ denotes the $i$th harmonic number. Since $H_i\sim\log i$ (natural logarithm), this yields $s(2^i)\sim\log i$ when $i\to\infty$. This suggests the asymptotics $$ \lim_{n\to\infty}\frac{s(n)}{\log\log_2 n}=\lim_{n\to\infty}\frac{s(n)}{\log\log n}=1. $$ Unfortunately, unless a more precise definition of $s(n)$ is provided, it seems that the sequence $(s(n))$ does not converge due to the jumps in the summation when $n$ passes a power of $2$. For example, for every $i$, $$ s(1+2^i)=\sum_{k=0}^{i}1/\log_2(2^{i-k}+2^{-k})\geqslant1/\log_2(1+2^{-i})\sim2^i\log2, $$ and $2^i\log2\gg\log\log(1+2^i)$. And actually, $$ \liminf_{n\to\infty}\frac{s(n)}{\log\log n}=1,\qquad\limsup_{n\to\infty}\frac{s(n)}{\log\log n}=+\infty. $$