Ʃ 1/√k ≥ 1/√n (under sigma should be "k=1" and above should be "n", and n is a positive integer)
The Attempt at a Solution
I. Base case when n=1 is correct.
II. Inductive Hypothesis: Assume true for k=m, where k<m<n, m is a positive integer.
III. Ʃ 1/√m + 1/√m+1 ≥ 1/√n, since Ʃ 1/√m ≥ 1/√n, and Ʃ 1/√m + 1/√m+1 ≥ Ʃ 1/√m.
By the principle of induction, Ʃ 1/√k ≥ 1/√n.
(I'm sorry for leaving out some subscripts under epsilon. I couldn't find how to get k=1 under there or "n" above).