show that harmonic series is divergent with $ \sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2} $ I'm pretty stuck with this exercise. Hope somebody can help me:
show that
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$
and use this to show that the harmonic series is divergent.
First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.

Writing out the series gave me this, but where do I go from here:
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$

Edit:
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.
 A: In fact
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}\ge \frac{1}{2^{k}+2^k}+\frac{1}{2^{k}+2^k}+\cdots+\frac{1}{2^{k}+2^k}=(2^{k+1}-2^k)\frac{1}{2^{k+1}}=\frac12.
$$
A: HINT
$$
\begin{split}
\sum_{n=2^k+1}^{2^{k+1}} \frac1n
 \ge \sum_{n=1}^{2^k} \frac{1}{2^k+n} 
 \ge \sum_{n=1}^{2^k} \frac{1}{2^{k+1}}
 = \frac{1}{2^{k+1}} \sum_{n=1}^{2^k} 1
\end{split}
$$
Can you finish?
A: (I am summarizing in this answer the steps discussed with OP in the comments)
Let $k\geq 0$ be an integer.
First note that the sum $\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}$ has $2^{k+1}-(2^{k}+1)+1= 2^{k+1}-2^k = 2^k$ terms, that is
$$\sum_{n=2^{k}+1}^{2^{k+1}} 1 = 2^k\tag{$*$}$$
Second, note that
$$\frac{1}{n}\geq \frac{1}{2^{k+1}},\qquad \forall n\, \text{ s.t. }\,2^{k}+1\leq n\leq 2^{k+1}\tag{$**$}$$
Combining the above observations, we have
$$\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}\,\overset{(**)}\geq\,\sum_{n=2^{k}+1}^{2^{k+1}}\frac{1}{2^{k+1}}\,=\,\frac{1}{2^{k+1}}\sum_{n=2^{k}+1}^{2^{k+1}}1\,\overset{(*)}{=}\, \frac{2^k}{2^{k+1}}\,=\,\frac{1}{2}\tag{$*\!*\!*$}$$
Now, note that
$$\sum_{n=1}^{2^{k+1}}\frac{1}{n}\,\geq\, \sum_{n=2}^{2^{k+1}}\frac{1}{n}\, =\, \sum_{m=0}^k\, \sum_{n=2^{m}+1}^{2^{m+1}}\frac{1}{n} \, \overset{(***)}{\geq}\, \sum_{m=0}^k \frac{1}{2}\, = \, \frac{k+1}{2}$$
As the above holds for every $k\geq 0$, by letting $k\to \infty$, we conclude that the harmonic series diverges.
