Sum of a divergent series 
For any $n \geq 5$, the value of $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n - 1}$ will lie between?

My attempt: I know the numbers are in Harmonic Progression but I don't know how to calculate $H_n$. However, I tried by approximately calculating the area under the curve of the subsequent rectangles formed $1\cdot1 + 1\cdot\frac{1}{2} + 1\cdot\frac{1}{3} + \cdots$ and so on) but I am still not any closer to the answer. Could anybody give some inputs? 
 A: You can compare the area of the susbequent rentangles with the area under the graph of function $f(x)=\frac{1}{x}$ to find following inequalities:
$$ \int_1^{2^n}\frac{1}{x}dx  < \sum_{k=1}^{2^n-1} \frac{1}{k} < 1 + \int_1^{2^n-1}\frac{1}{x}dx$$
That is
$$ n \ln 2   < \sum_{k=1}^{2^n-1} \frac{1}{k} < 1 + \ln(2^n-1) = n\ln 2 + 1 + \ln(1-2^{-n})$$
Which gives an approximation that is good enough for most purposes.
A: Note that 
$$\sum^{2^{n}-1}_{i=1}\frac{1}{i}=\sum_{k=1}^{n}\sum^{2^{k}-1}_{i=2^{k-1}}\frac{1}{i}$$
And thus we find
$$\sum_{k=1}^{n}\sum^{2^{k}-1}_{i=2^{k-1}}\frac{1}{i}\leq\sum_{k=1}^{n}\sum^{2^{k}-1}_{i=2^{k-1}}\frac{1}{2^{k-1}}=\sum_{k=1}^{n}2^{k-1}\frac{1}{2^{k-1}}=n.$$
Also
$$\sum_{k=1}^{n}\sum^{2^{k}-1}_{i=2^{k-1}}\frac{1}{i}\geq\sum_{k=1}^{n}\sum^{2^{k}-1}_{i=2^{k-1}}\frac{1}{2^{k}}=\sum_{k=1}^{n}2^{k-1}\frac{1}{2^{k}}=\frac{1}{2}n.$$
Hence $\frac{1}{2}n\leq H_{n}\leq n.$$
Obviously, there are sharper bounds, but without any context I'm not sure how sharp you need them to be.
