TITLE: Does this series diverge: $(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)+\dots$?
QUESTION [6 upvotes]: $$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$
I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.
Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?

REPLY [2 votes]: @Jasper Joy is indeed right. It's a telescoping series whose partial sums eventually only have a fixed number of terms after cancellation:
$$(\sqrt{n+1} - \sqrt n) + ... + (\sqrt 4 - \sqrt 3) + (\sqrt 3 - \sqrt 2) + (\sqrt 2 - \sqrt 1)$$
$$= \sqrt{n+1} + (-\sqrt n + \sqrt n) + ... + (-\sqrt 3 + \sqrt 3) + (-\sqrt 2 + \sqrt 2) - \sqrt 1$$
$$= \sqrt{n+1} - 1$$
In series sigma notation we can summarize this as:
$$\sum_{i=1}^{n} (\sqrt{i+1}-\sqrt{i}) = {\sqrt{n+1} - 1}$$
As an infite sum it diverges to $\infty$:
$$\sum_{i=1}^{\infty} (\sqrt{i+1}-\sqrt{i}) = \lim_{i \to \infty}{\sqrt{i+1} - 1}$$