Question

Prove that the mathematical induction that
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...,\frac{1}{2^n}\ge 1+\frac{n}{2}$
for each no negative interger n

Answer 1

p(0) , p(1) , clearly holds goods
$p\left(2\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}>\frac{24}{12}=21+\frac{2}{2}$
Thus p(2) holds goods ,
p(n + 1) = $(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}......+\frac{1}{2^n})+\frac{1}{2^{n+1}}$
=p(n) + $(\frac{1}{1+2^n}+\frac{1}{2+2^n}+\frac{1}{2^n+2^n})$
The 2 nd bracket contain 1 , 2 , 3 , .........$2^n$ . i.e. 2 terms Each term of this bracket
$>\frac{1}{2^n+2^n}+\frac{1}{2\cdot 2^n}\frac{1}{2^{n=1}}$
$\therefore p(n+1)\ge (1+\frac{n}{2})+2^n\cdot \frac{1}{2^{n+1}}$
$=1+\frac{n}{2}+\frac{1}{2}=1+\frac{n+1}{2}$
$\therefore p(n+1)\ge 1+\frac{n+1}{2}$
Thus p(n ) is universally true