Answer to Question #217592 in Discrete Mathematics for Raja

Question

For any natural number n, prove the validity of given series by mathematical induction:

2(√(n+1)-1)<1+(1/√2)+⋯..+(1/√n)<2√n?

Answer 1

$2\left( \sqrt{n+1}-1 \right)<1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{n}}<2\sqrt{n}$

Base Case: $n=1$

$2\left(\sqrt{2}-1\right)<1<2$

$\sqrt{2}-1<\tfrac{1}{2}<1$

$\sqrt{2}\approx 1.414<1.5<2$

Assume: $2\left( \sqrt{k+1}-1 \right)<1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k}}<2\sqrt{k}$

Prove: $2\left( \sqrt{k+2}-1 \right)<1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k+1}}<2\sqrt{k+1}$

1) $1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k}}+\tfrac{1}{\sqrt{k+1}}<2\sqrt{k}+\tfrac{1}{\sqrt{k+1}}=\frac{2\sqrt{k(k+1)}+1}{\sqrt{k+1}}$

Using inequality $\sqrt{xy}\leq \frac{x+y}{2}$ , we get $\frac{2\sqrt{k(k+1)}+1}{\sqrt{k+1}}\leq \frac{(2k+1)+1}{\sqrt{k+1}}=2\sqrt{k+1}$

Therefore, $1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k+1}}<2\sqrt{k+1}$

2) $1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k}}+\tfrac{1}{\sqrt{k+1}}>2\left(\sqrt{k+1}-1\right)+\tfrac{1}{\sqrt{k+1}}=\frac{2(k+1)+1}{\sqrt{k+1}}-2=\frac{2k+3}{\sqrt{k+1}}-2$

Using inequality $\sqrt{xy}\leq \frac{x+y}{2}$ , we get $\sqrt{(k+1)(k+2)}\leq \frac{2k+3}{2}$ .

So, $2\sqrt{k+2}\leq \frac{2k+3}{\sqrt{k+1}}$ .

Therefore, $1+\tfrac{1}{\sqrt{2}}+…+\tfrac{1}{\sqrt{k+1}}>2\sqrt{k+2}-2=2\left(\sqrt{k+2}-1\right)$