Use the mathematical Induction show that $H_{2^n}\le n+1$
here $H$ is harmonic numbers ie. $H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}$
my idea
so for $n=0$ L.H.S=R.H.S
Suppose this is true for $n$
we prove for $n+1$
So $H_{2^{n+1}}=1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{2^n}+\frac{1}{2^n+1}+....\frac{1}{2^{n+1}}\\ =H_{2^n}+\frac{1}{2^n+1}+.......\frac{1}{2^{n+1}}$
How to continue from here
We have already assumed that $$H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}\le n+1$$
We need to prove that:
$$H_{n+1}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n}+\frac{1}{2^n+1}+...+\frac{1}{2^n+2^n}\le n+2$$
This is true because $$\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n}<\frac{1}{2^n+1}+\frac{1}{2^n+1}+...+\frac{1}{2^n+1}\text{(there are $2^n$ numbers)}$$
$$\Rightarrow \frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n}<\frac{2^n}{2^n+1}<1$$
Combined with: $$H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}\le n+1$$ to finish the proof.
Rephrasing a bit:
1) $n=0$, ok.
2) Hypothesis: $H_{2^n} \le n+1$.
3)Step: $n+1$:
$H_{2^{n+1}} =$
$H_{2^n} + \dfrac{1}{2^n +1}+.....\dfrac{1}{2^{n+1}} =$
$H_{2^n} + \dfrac{1}{2^n+1}+...\dfrac{1}{2^n+ 2^n} \lt$
$H_{2^n} + 2^n \dfrac{1}{2^n+1} \lt $
$H_{2^n} +1 \lt (n+1) +1$, where
in the second last line $\dfrac{2^n}{2^n+1} <1$,
and in the last line the hypothesis has been used.
$$H_n = \sum_{k=1}^n \frac{1}{k}.$$ Then $$\begin{align*} H_{2^{n+1}} &= \sum_{k=1}^{2^{n+1}} \frac{1}{k} \\ &= \sum_{k=1}^{2^n} \frac{1}{k} + \sum_{k=2^n + 1}^{2^{n+1}} \frac{1}{k} \\ &= H_{2^n} + \sum_{j=1}^{2^{n+1} - 2^n} \frac{1}{2^n + j} \\ &\overset{\text{i.h.}}{\le} (n+1) + \sum_{j=1}^{2^n} \frac{1}{2^n + j} \\ &< (n+1) + \sum_{j=1}^{2^n} \frac{1}{2^n} \\ &= (n+1) + 2^n \cdot \frac{1}{2^n} \\ &= (n+1) + 1 \\ &= n+2, \end{align*}$$ where the inequality with "i.h." indicates the step where the induction hypothesis was used.
