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?