convergence of $\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}$

Question

I need to determine if the series $$\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}$$ converges or diverges.

My work so far:

Using the comparison test, $\frac{1}{\sqrt n + \sqrt{n+1}} < \frac{1}{\sqrt n } $ and $\frac {1}{\sqrt n }$ is a divergent series

Answer 1

There is no point of dominating a given sequence by a divergent sequence to show divergence. Conclusions can be drawn in the following cases:

1) The function dominates (loosely, is greater than) a divergent sequence, in which case it is divergent.

2) The function is dominated by (loosely, is smaller than) a convergent sequence, in which case it is convergent.

This question should be done as follows: Note that $\frac{1}{\sqrt n + \sqrt{n+1}} = \frac{\sqrt {n+1} - \sqrt n}{(n+1)-n} = \sqrt{n+1}-\sqrt{n}$, and that the sum $\sum \sqrt{n+1}-\sqrt n$ is divergent, hence the original sum is divergent.

Answer 2

It makes no sense to bound above by a divergent series. Instead, use this:

$$\frac1{\sqrt n+\sqrt{n+1}}>\frac1{\sqrt{n+1}+\sqrt{n+1}}=\frac1{2\sqrt{n+1}}$$

which is then appropriate to show divergence since it is a lower bound.