finding partial sum of series with square root

Question

I'm trying to practice finding formulas of sums (partial sums). stuck a bit with this sum: $$ \sum_{n=1}^n\frac{1}{\sqrt{n+1}+\sqrt{n}} $$

the answare should be $$S_n=\sqrt{n+1}-1$$

Answer 1

Rewriting the sum in a more managable way we have: \begin{equation} \sum_{n=1}^k\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\sum_{n=1}^k\dfrac{1}{\sqrt{n+1}+\sqrt{n}} \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\sum_{n=1}^k{\sqrt{n+1}-\sqrt{n}} \end{equation} Now: \begin{equation} \sum_{n=1}^k{\sqrt{n+1}-\sqrt{n}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{k+1}-\sqrt{k}= \sqrt{k+1}-1 \end{equation} The series does not converge as $\displaystyle{\lim_{k \to +\infty} S_k= +\infty}$. You can see that it diverges also using asymptotic behaviour: \begin{equation} \sum_{n=1}^{+\infty}\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\sim \sum_{n=1}^{+\infty}\dfrac{1}{\sqrt{n}} \end{equation}