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The harmonic series, $\sum_{k=1}^{\infty}\frac{1}{k}$, diverge. Let the partial sum be denoted as $s_n = \sum_{k=1}^{n}\frac{1}{k}$. If $s_n$ was convergent, then it would have to be bounded and its subsequence $s_{2^n}$ must be also bounded. But, $$s_{2^n} = (1) + (\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+\dots+(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n})>1+\frac{n}{2}$$ as $n \rightarrow \infty$. I can see that each term in parenthesis are greater than $\frac{1}{2}$. But the book doesn't show how $(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}) > \frac{1}{2}$ for each $n$. Is this supposed to be intuitively obvious or is there some way to prove this rigorously?

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    In each parenthesized group, replace each fraction by the smallest one, and add the new values. – 2017-01-26
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    @quasi Thank you! I see it now – 2017-01-26
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    There are $2^{n-1}$ terms each greater or equal to $1/2^n.$ – 2017-01-26

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$$\left(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}\right) > \underbrace{\left(\frac{1}{2^n}+\dots+\frac{1}{2^n}\right)}_{2^{n-1} \text{times}} = \frac{2^{n-1}}{2^n} = \frac{1}{2} $$

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    How do you know there are $2^{n-1}$ of $\frac{1}{2^n}$? – 2017-01-26
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    $2^n-2^{n-1}=2^{n-1}$. – 2017-01-27
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Another way

Let $\sum_{k=1}^{\infty}\frac 1k=L$ where $L\in \mathbb{R}$. We have $$L=1+\frac 12+\frac 13+\frac 14+\frac 15+\frac 16+\cdots$$ $$L>\frac 12+\frac 12+\frac 14+\frac 14+\frac 16+\frac 16+\cdots$$ $$L>1+\frac 12+\frac 13+\cdots$$ $$L>L$$ It is contradiction.

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    There is a problem with this argument. You can assert $L = \lim_{N \to \infty} \sum_{k=1}^N (1/2^{k-1} + 1/2^k)$ and $ \sum_{k=1}^N (1/2^{k-1} + 1/2^k) > \sum_{k=1}^N 2/2^k = \sum_{k=1}^{N-1} 1/k$. However taking the limit as $N \to \infty$ you can only conclude $L \geqslant \sum_{k=1}^\infty 1/k$. Note the inequality need not be strict given only that the RHS sequence is convergent and termwise strictly less than $L$. For example $1-1/n < 1$ but $1-1/n \to 1$. – 2017-01-26