I'm trying to understand the following proof for the divergence of the harmonic series:
The parts which I can't seem to understand are:
$$1. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{r} > \sum_{i=0}^{n-1}
\sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}}$$
$$2. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}} = \sum_{i=0}^{n-1} (2^{i+1} – 2^i)\frac{1}{2^{i+1}}$$
I think understanding 2. follows from understanding 1. but I'm not entirely sure. Anyway for 1. I couldn't understand if the series on the RHS of the equality was $(\frac{1}{2^3})+(\frac{1}{2^3} + \frac{1}{2^4})+(\frac{1}{2^5} + \frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8})+…$ or $(\frac{1}{2^1})+(\frac{1}{2^2} + \frac{1}{2^3})+(\frac{1}{2^4} + \frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7})+…$ . And for part I simply could not understand 2. (probably because I didn't understand 1.). So I'm wondering what the series on the RHS of 1. actually is and how it equals $\sum_{i=0}^{n-1} (2^{i+1} – 2^i)\frac{1}{2^{i+1}}$ .
Best Answer
It's useful to look just at the difference between each term in the sequence $\{S_{2^n}\}$, i.e., $$S_{2^{n+1}} - S_{2^n} = \sum_{k = 2^n+1}^{2^{n+1}} \frac{1}{k} \geq \sum_{k = 2^n+1}^{2^{n+1}} \frac{1}{2^{n+1}} = \frac{1}{2^{n+1}} \sum_{k = 2^n+1}^{2^{n+1}} 1 = \frac{2^n}{2^{n+1}} = \frac{1}{2}.$$
So $S_{2^{n+1}}\geq S_{2^n} + \frac{1}{2} $ for all $n$. If the partial sums increase by at least $\frac{1}{2}$ each time, the series must diverge to infinity.