68. ∑(√(k + 1) - √k) from k = 1 to ∞

Explanation:

Step1: Expand the partial - sum

Let (S_N=\sum_{k = 1}^{N}(\sqrt{k + 1}-\sqrt{k})). Expand the sum: [ \begin{align*} S_N&=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{N + 1}-\sqrt{N}) \end{align*} ]

Step2: Simplify the sum by cancellation

By cancellation, most of the terms will be eliminated. We have (S_N=\sqrt{N + 1}-1).

Step3: Find the limit as (N\to\infty)

We want to find (\lim_{N\to\infty}S_N=\lim_{N\to\infty}(\sqrt{N + 1}-1)). As (N\to\infty), (\sqrt{N+1}\to\infty), so (\lim_{N\to\infty}(\sqrt{N + 1}-1)=\infty).

Answer:

The series (\sum_{k = 1}^{\infty}(\sqrt{k + 1}-\sqrt{k})) diverges.