Prove that $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \mathcal{O}(\log(n)) $.

\begin{align} \sum_{k=1}^n\dfrac{1}{n}&\leq1+ \int_1^n\dfrac{1}{x}dx\\&=1+\log(x)\Big|_1^n\\&=1+\log(n)-\log(1)\\&=1+\log(n)\\&=\mathcal{O}(\log(n)) \end{align}

Using the inequality $1+x\le e^x$, we can derive $$ \log\left(\frac{n+1}{n}\right)=\log\left(1+\frac1n\right)\le\frac1n\le-\log\left(1-\frac1n\right)=\log\left(\frac{n}{n-1}\right)\tag{1} $$ Summing $(1)$ yields $$ \log(n+1)\le\sum_{k=1}^{n}\frac1k=1+\sum_{k=2}^{n}\frac1k\le1+\log(n)\tag{2} $$ That is, for all $n\ge1$ $$ \log(n+1)\le\sum_{k=1}^{n}\frac1k\le1+\log(n)\tag{3} $$

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Sums can easily be made into inductions. We will prove $(3)$ by induction using $(1)$.

For $n=1$, $(3)$ holds since $\log(2)\le1\le1$.

Suppose we have $(3)$ for $n-1$: $$ \log(n)\le\sum_{k=1}^{n-1}\frac1k\le1+\log(n-1)\tag{4} $$ Inequality $(1)$ says that $$ \log\left(\frac{n+1}{n}\right)\le\frac1n\le\log\left(\frac{n}{n-1}\right)\tag{5} $$ Adding $(4)$ to $(5)$ yields $$ \log(n+1)\le\sum_{k=1}^{n}\frac1k\le1+\log(n)\tag{6} $$ which is $(3)$ for $n$. This finishes the induction.

I'm expanding the answer by xan:

Define $H_n=\displaystyle\sum_{1\le k\le n} {1\over k}$, let's prove by induction that $H_{2^n}\le n+1$. This is true for $n=0$ since $H_{2^0}=H_1=1\le 1$.

Now suppose $H_{2^n}\le n+1$. We have:

$$\begin{align} H_{2^{n+1}} &= \sum_{1\le k\le 2^{n+1}} {1\over k} \\ H_{2^{n+1}} &= \sum_{1\le k\le 2^n} {1\over k} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over k} \\ H_{2^{n+1}} &= H_{2^n} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over k} \\ H_{2^{n+1}} &\le H_{2^n} + \sum_{2^n+1\le k\le 2^{n+1}} {1\over 2^n} \\ H_{2^{n+1}} &\le H_{2^n} + (2^n-1){1\over 2^n} \\ H_{2^{n+1}} &\le H_{2^n} + (1-{1\over 2^n}) \\ H_{2^{n+1}} &\le H_{2^n} + 1 \\ H_{2^{n+1}} &\le (n+1) + 1 \\ H_{2^{n+1}} &\le n+2 \\ \end{align} $$

Now let's make $m=2^n$, then $n=\lg m$, and:

$$H_{2^n}=H_m\le\lg m+1=\mathcal{O}(\log m)$$