Tricky sequences and series problem For a positive integer $n$, let $a_{n}=\sum\limits_{i=1}^{n}\frac{1}{2^{i}-1}$. Then are the following true:
$a_{100} > 200$ and 
$a_{200} > 100$?
Any help would be thoroughly appreciated. This is a very difficult problem  for me. :(

Clearly, for $k>1$,
$$
\int_k^{k+1}\frac{dx}{x}<\frac{1}{k}<\int_{k-1}^{k}\frac{dx}{x}.
$$
Hence
$$
n\log 2=\log (2^n)=\int_1^{2^n}\frac{dx}{x}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n-1}<1+\int_1^{2^n-1}\frac{dx}{x}=1+\log(2^n-1),
$$
and thus
$$
n\log 2<a_n<1+\log(2^n-1)<1+n\log 2.
$$
So
$$
a(100)<1+100\log 2<1+100=101,
$$
as $\log 2<\log e=1$, and
$$
a(200)>200\log 2=100 \log 4>100,
$$
as $\log 4>\log e=1$.
