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. :(
 A: $$\begin{gather}a_n = 1 + \underbrace {\frac{1}{2} + \frac{1}{3}}_{2\text{ terms}} + \underbrace {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}}_{4\text{ terms}} + \ldots + \underbrace { \frac{1}{2^{n -1}} + \frac{1}{2^{n -1}+1} + \ldots +  \frac{1}{2^n -1}}_{2^{n-1}\text{ terms}}\\
> 1 + 2\cdot\frac{1}{4} + 4\cdot\frac{1}{8} + \ldots + 2^{n-1}\cdot\frac{1}{2^{n}}
\\ =1 + \underbrace { \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2}}_{(n-1) \text{ terms}} = 1 +\frac{n-1}{2} = \frac{n+1}{2}.
\end{gather}
$$
Thus, $a_n > \frac{n+1}{2}.$
On the other hand,
$$a_n < 1 + 2\cdot\frac{1}{2} + 4\cdot\frac{1}{4} + \ldots + 2^{n-1}\cdot\frac{1}{2^{n-1}} = 1+(n-1 )= n,$$
so the inequality $a_{100} > 200$ cannot be true.
A: for $a_{100}>200$ I think you can say $200=\sum\limits_{i=1}^{100}2$ then,
$a_{100}-200=\sum\limits_{i=1}^{100}\frac{1}{2^{i}-1}-2=\sum\limits_{i=1}^{100}
\frac{3-2^{i+1}}{2^{i}-1}$ and this sum is clearly negative. So $a_{100}>200$ is wrong.
For $a_{200}>100$ you can say $100=\sum\limits_{i=1}^{200}\frac{1}{2}$ and do the same argument. 
I hope this will be helpful. 
A: 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 $\frac{1}{5} < \frac{1}{4}, \frac{1}{6} < \frac{1}{4}...$
A: $a(100)>200$ is not true; $a(200)>200$ is true.
We need the following fact:

$\frac{1}{1+n}<\ln (1+\frac{1}{n}) <\frac{1}{n}.$

$a(100)<1+\ln 2+\ln\frac32+\ln\frac43+\dots+\ln\frac{2^{100}}{2^{100}-1}=1+100\ln2<101<200.$
$a(200)>\ln2+\ln\frac32+\dots+\ln\frac{2^{200}+1}{2^200}=\ln(2^{200}+1)>\ln 4^{100}>100\ln 4>100.$
A: 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 $\log 4>\log e=1$.
