TITLE: Sigma index (Induction)
QUESTION [5 upvotes]: I was wondering how the index that's left is measured.
When you prove this via induction,
$$\sum_{k=1}^{2^n} \frac{1}{k} \geq \frac{n}{2} $$
$$\sum_{k=1}^{2^{n+1}} \frac{1}{k}= \sum_{k=1}^{2^{n}} \frac{1}{k} + \sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k} $$
you will come across this part
$$\geq\frac{n}{2}+\sum_{k=2^n+ 1}^{2^{n+1}} \frac{1}{k}\geq \frac{n}{2}+2^n\cdot \frac{1}{2^{n+1}}$$
where does the  $2^n$ comes from?
Is it a index difference?

REPLY [2 votes]: Yes, it depends on the limits of the sum. More generally, if $(x_k)_{k\geq 1}$ is a decreasing sequence then, for $b\geq a\geq 1$,
$$\sum_{k=a+1}^bx_k\geq \sum_{k=a+1}^bx_b=x_b\sum_{k=a+1}^b1=(b-a)\cdot x_b.$$
In your case, $x_k=1/k$, $b=2^{n+1}$, and $a=2^n$.
Therefore
$$(b-a)=2^{n+1}-2^n=2^n.$$