Proving 1 + 1/√2 +...+ 1/√n < 2√n Using Mathematical Induction

[trap101]

Prove that:

1 + 1/√2 +...+ 1/√n < 2√n

Work:

So I've done the base case of n = 1, and I've set up the Indutive hypothesis as assuming n=k as 1 + 1/√2 +...+ 1/√k < 2√k

and for the inductive step:

1 + 1/√2 +...+ 1/√k + 1/√(k+1) < 2√k + 1/√(k+1)

now I'm having trouble trying to manipulate 2√k + 1/√(k+1) to get 2√(k+1) alone. I tried a conjugate of 1/[2 √(k+1) + 2√k] [2 √(k+1) - 2√k / [2 √(k+1) - 2√k]...the closest thing I got was 2√(k+1) / 4.

ANy suggestions?

 

[jbunniii]

Here is a hint which should be useful:
$$4(k^2 + k) < 4k^2 + 4k + 1 = (2k + 1)^2$$
Taking the square root of both sides, which is OK because both sides are positive, we get
$$2\sqrt{k^2 + k} < 2k + 1$$
See if you can manipulate the right hand side of your inequality into a form where you can apply this.

 

[Ray Vickson]

Prove that:

1 + 1/√2 +...+ 1/√n < 2√n

Work:

So I've done the base case of n = 1, and I've set up the Indutive hypothesis as assuming n=k as 1 + 1/√2 +...+ 1/√k < 2√k

and for the inductive step:

1 + 1/√2 +...+ 1/√k + 1/√(k+1) < 2√k + 1/√(k+1)

now I'm having trouble trying to manipulate 2√k + 1/√(k+1) to get 2√(k+1) alone. I tried a conjugate of 1/[2 √(k+1) + 2√k] [2 √(k+1) - 2√k / [2 √(k+1) - 2√k]...the closest thing I got was 2√(k+1) / 4.

ANy suggestions?

So, you want to show $$2 \sqrt{k+1} - 2 \sqrt{k} \geq \frac{1}{\sqrt{k+1}}.$$ Re-write the LHJS using the type of trick you applied above.