Mathematical Induction Problem

[klarka]

Homework Statement

Prove that $$\frac{1}{\sqrt{1}}$$+$$\frac{1}{\sqrt{2}}$$+...+$$\frac{1}{\sqrt{n}}$$$$\geq$$$$\sqrt{n}$$ for all n $$\in$$ N

Homework Equations

The Attempt at a Solution

p(1): $$\frac{1}{\sqrt{1}}$$ = $$\frac{1}{1}$$ = 1 = $$\sqrt{1}$$ $$\geq$$ $$\sqrt{1}$$

Let $$\frac{1}{\sqrt{1}}$$ + $$\frac{1}{\sqrt{2}}$$ +...+ $$\frac{1}{\sqrt{n}}$$ $$\geq$$ $$\sqrt{n}$$ for some n$$\in$$ N

1/√1 + 1/√2 + ... + 1/√(n+1) > √n + 1/√(n+1)

= (√n√(n+1) + 1)/√(n+1)

=(√n(n+1) + 1)/√(n+1)

 

[Mentallic]

For an induction problem, after your assumption for some $$n=k, n\in N$$ you're supposed to prove the condition is true for $$n=k+1$$, thus

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\geq\sqrt{k+1}$$

You can start by taking the last term $$\frac{1}{\sqrt{k+1}}$$ onto the right side. Now that you've assumed that first expression, if you can prove $$\sqrt{k}\geq \sqrt{k+1}-\frac{1}{\sqrt{k+1}}$$ then you have essentially finished the proof. Do you understand why?