- #1

trap101

- 342

- 0

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?