Prove that $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \mathcal{O}(\log(n)) $. Prove that $ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} = \mathcal{O}(\log(n)) $, with induction.
I get the intuition behind this question. Clearly, the given function isn’t even growing at a linear rate, but what is the ‘proper’ proofy way to say that $ \displaystyle \sum_{k=1}^{n} \frac{1}{k} \leq \mathcal{O}(\log(n)) $? I was unable to find any useful identities to use for such a summation.
 A: 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 $n\ge1$
$$
\log(n+1)\le\sum_{k=1}^{n}\frac1k\le1+\log(n)\tag{3}
$$

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.
A: 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:
$$H_{2^n}=H_m\le\lg m+1=\mathcal{O}(\log m)$$
A: Hint: Think about the integral $\displaystyle\int_1^n \frac{dx}x$.
A: \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}
A: You want to show that
$\sum_{k=1}^{n} \frac{1}{k} \leq \mathcal{O}(\log(n))$.
This means that there is a constant $c > 0$
and an integer $n_0$
such that
$\sum_{k=1}^{n} \frac{1}{k} \leq c\log(n)$
for all $n \ge n_0$.
Suppose that this is true for some $n$ and $c$.
We want to show that
$\sum_{k=1}^{n+1} \frac{1}{k} \leq c\log(n+1)$
follows.
By the induction hypothesis,
$\sum_{k=1}^{n+1} \frac{1}{k}
= \sum_{k=1}^{n} \frac{1}{k} + \frac1{n+1}
\leq c \log(n) + \frac1{n+1}
$
.
If we could show that
$c \log(n) + \frac1{n+1} \le c \log(n+1)$,
we would be done.
But $\log(n+1) - \log(n)
= \log(1+1/n) > 1/n
> 1/(n+1)
$,
so this inequality holds
for any $c \ge 1$.
To find a particular $c$ and $n_0$,
look at $n = 3$.
$1 + 1/2 + 1/3 < 2 <
c \log 3$ for $c = 2$.
So $c=2$ will work.
Of course the best value of $c$ is 1,
but that is not needed for a $\mathcal{O}$
result - there just needs to be some $c$.
