# Series convergence (or divergence)

I've got the following exercise:
Show that the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}+\sqrt{n+1}}$ converges and compute its limit if it's posible.
I've already tried with the Integral test but is worthless, I've got the same thought about using the ratio test. So I don't know which one of the convergence test would be most useful to compute the limit.

• Compare to $\sum_{n \geq 1} \frac{1}{\sqrt{n+1} + \sqrt{n+1}} = \infty$, by the integral test. – Dzoooks Jun 5 '18 at 2:11
• $>\frac{1}{2\sqrt{n+1}}>\frac{1}{2(n+1)}$ – W. mu Jun 5 '18 at 2:11
• I'd be shocked if it converges. – Randall Jun 5 '18 at 2:11
• I know it's not the problem, but an alternative interesting question would be to replace the $+$ between the radicals with a $-$. That one also has an answer... – imranfat Jun 5 '18 at 2:15

hint: Use comparison test with the series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}$, and this one is for sure divergent. Or use the definition of the partial sum $S_n = \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{k}+\sqrt{k+1}} = \displaystyle \sum_{k=1}^n \left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - 1\to \infty$ for $n \to \infty$ .