Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$ 
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
  
  
*
  
*(a) $S(100)\leq 100$.
  
*(b) $S(100)>100$.
  
*(c) $S(200)\leq 100$.
  
*(d) $S(200)>100$.
  

My attempt


*

*For the upper bound 
$$\begin{align}
S(n) &= 1 + \left( \frac 12 + \frac 13 \right) + \left( \frac 14 + \frac 15 + \frac 16 + \frac 17 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}+1} + \cdots + \frac 1{2^n-1} \right)
\\ &< 1 + \left( \frac 12 + \frac 12 \right) + \left( \frac 14 + \frac 14 + \frac 14 + \frac 14 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}} + \cdots + \frac 1{2^{n-1}} \right) \\
&= \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} \\
&= n.
\end{align}$$
So we get $S(n) < n$ (for $n > 1$), and in particular $S(100) < 100$.
Now I did not understand how to calculate a lower bound, or if there is any other method by which we can solve this.

This is not an answer but it is too long for a comment.
Since $$S(n)=\sum_{i=1}^{2^n-1}\frac 1i=H_{2^n-1}$$ you could be interested by this paper in which the author, Mark B. Villarino, developed extremely sharp bounds for the harmonic numbers.
The first and simplest set of bounds (Theorem $1$) is given by $$\frac{1}{2 m+\frac{1}{1-\gamma }-2}+\log (m)+\gamma\leq H_m \lt \frac{1}{2 m+\frac{1}{3}}+\log (m)+\gamma$$
Applied to the cases given by Dr. MV, the bounds coincide to more than $50$ significant figures.
