What is $\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$? I am stuck on this one; Its a sum and don't know how to calculate the denominator.
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$$

Stolz-Cesaro lemma is applied twice successively:
$$L=\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}} =\lim_{n\to\infty} \frac{1}{n+2-(n+1)} \frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\lim_{n\to\infty}\frac{n+1}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2}}=\frac{n+2-(n+1)}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+3}-(1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{n+2})}=\lim_{n\to\infty}\frac{1}{\sqrt{n+3}}=0$$
