Proof of sequence inequality

Question

I am wondering how to prove both of this by induction: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^n}\leq 1+\frac{n}{2}$ In general, how should I think of how to prove such an inductive step for these types of inequalities? If there is any other way to prove please feel free. Any help would be appreciated.

Answer 1

Induction

When $n=1$, $1+\frac{1}{2}\geq 1+\frac{1}{2}$, the inequality holds.

Suppose the inequality holds when $n=k$, where $k \in \mathbb{N}$. $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^k}\geq 1+\frac{k}{2}$$

When $n=k+1$,

$LHS=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^k})+(\frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots+\frac{1}{2^{k+1}})\\ \geq(1+\frac{k}{2})+(\frac{1}{2^{k+1}}\cdot2^k)=1+\frac{k+1}{2}=RHS$

The inequality also holds.

Another Solution

$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots+\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\cdots+\frac{1}{2^n}\\ \geq 1+ \frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots+\frac{1}{2^n}+\frac{1}{2^n}+\cdots+\frac{1}{2^n}\\ =1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2}\\ =1+\frac{n}{2}$

Answer 2

$\sum\limits_{n=2^{i}+1}^{2^{i+1}}\frac{1}{n}\geq \sum\limits_{n=2^{i}+1}^{2^{i+1}}\frac{1}{2^{i+1}}=\frac{1}{2}$, therefore $\sum\limits_{n=1}^{2^k}\frac{1}{n}=1+\sum\limits_{i=0}^{k-1}\sum\limits_{j=2^{i}+1}^{2^{i+1}}\frac{1}{j}\geq 1+\frac{k}{2}$.

Answer 3

Another solution using harmonic numbers

$$\sum_{i=1}^{2^n}\frac 1 i=H_{2^n}$$

Using the asymptotics, $$H_{2^n}=\gamma + n \log(2)+\frac 1 {2^{n+1 }}-\frac 13 \frac 1 {2^{2(n+1) }}+\cdots$$ $\log(2) > \frac 12$ so, as soon as $n>2$ the inequality holds.

For $n=2$ the lhs is $\frac{25}{12}=2+\frac 1 {12}$ while the rhs is $2$.