Evaluate $\sum \sqrt{n+1} - \sqrt n$

Question

$$\sum_{n=1}^\infty (\sqrt{n+1} - \sqrt n).$$

so for this, I multiplied by the conjugate $\sqrt {n+1}+ \sqrt n$ and then got limit $1/(n+1-n^2)$ however says it converges to zero?

Answer 1

It is easy to see that the sum does not converge.

Imagine you add only the first $k$ terms of the series. You got

$$ (\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{k}-\sqrt{k-1})+(\sqrt{k+1}-\sqrt{k}) $$

Now you can perform the addition, and you see that all the terms, apart the first ($-\sqrt{1}$) and the very last $\sqrt{k+1}$ cancel out.

So the sum of the first $k$ terms of the series is just $\sqrt{k+1}-1$, thus the series cannot converge.

This technique of telescoping is useful when your series can be put in the form $$ \sum_{i=1}^{\infty}(f(i+1)-f(i)) $$ or something similar.

Your approach of multiplying by the conjugate can be done, but you did in the wrong way.

You should multiply and divide for the same quantity, otherwise you are changing the value of the terms of the serie!

Here

$$ \frac{\sqrt{n+1}-\sqrt{n}}{1}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}} $$

So you have to compute $$ \sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}+\sqrt{n}} $$

Again it is easy to see that this cannot converge because $$ \frac{1}{\sqrt{n+1}+\sqrt{n}}>\frac{1}{2\sqrt{n+1}}>\frac{1}{n} $$ for $n>4$, because for $n>4$ we have $n>2\sqrt{n+1}$. And I assume you know that the series $\sum\frac{1}{n}$ does not converge.