<s>[INST] 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. :(
 [/INST] Have you thought about regrouping some of the terms ?
$\frac{1}{2} \geq \frac{1}{2}$
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$ since $\frac{1}{3} > \frac{1}{4}$
$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{2}$ since $\frac{1}{5} > \frac{1}{8}, \frac{1}{6} > \frac{1}{8}...$
...
$\frac{1}{2^{199}+1} + ... +\frac{1}{2^{200}-1} \geq \frac{1}{2} - \frac{1}{2^{200}}$
Finally, we don't forget the $1$ :
$1 = 2*\frac{1}{2}$.
If we sum everything, we get : $1+ \frac{1}{2}+ \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{2^{200}-1} > 201*\frac{1}{2} - \frac{1}{2^{200}} > 100$.
So the answer to your second question is YES.
The other answer is NO : you can prove it by regrouping the terms differently, with inequalities the other way around :
$\frac{1}{2} + \frac{1}{3} < 1$ since $\frac{1}{3} < \frac{1}{2}$
$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < 1$ since $\frac{1}{5} < \frac{1}{4}, \frac{1}{6} < \frac{1}{4}...$
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