Prove that ∑_(k=1)^∞ 1/k is divergent
Could anyone possibly help explain my professor’s notation in this proof? Struggling to understand where S_(2^n-1)≥n/2 came from as well as the manipulation of the sums that follow.
Any help is greatly appreciated
On
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+\frac{1}{2} + ....+ \frac{1}{n} $. Then
$$(*) \quad S_{2n}= S_n +\frac{1}{n+1} +...+ \frac{1}{2n} \ge S_n +\frac{1}{2n} +...+ \frac{1}{2n}= S_n +\frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S \ge S+\frac{1}{2}$, a contradiction.
Any sum of the form $$\sum_{i=1}^{2^n-1}f(i)$$ can be written as the double sum $$\sum_{k=1}^n\sum_{i=2^{k-1}}^{2^k-1}f(i)$$ To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that $$\sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$ $$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$ $$=\sum_{i=2^{1-1}}^{2^1-1}f(i)+\sum_{i=2^{2-1}}^{2^2-1}f(i)+\sum_{i=2^{3-1}}^{2^3-1}f(i)$$ $$=\sum_{k=1}^3\sum_{i=2^{k-1}}^{2^k-1}f(i)$$ See it now? Hence we can write $$S_{2^n-1}=\sum_{k=1}^n\sum_{i=2^{k-1}}^{2^k-1}\frac{1}{i}=1+\sum_{k=2}^n\sum_{i=2^{k-1}}^{2^k-1}\frac{1}{i}$$ and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.