Limit of a certain quotient $$ \lim_{n\to \infty} {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}} $$
I understand that the summation is divergent and this is $\infty\over\infty$ form. But how to proceed further??
 A: Using Stolz-Cesaro theorem (which is pretty useful) we get:
$\displaystyle\lim_{n\to \infty} {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}}=\lim_{n\to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=2$
A: $$\lim_{n\to \infty}\frac{\sum_{k=1}^{n}\frac{1}{\sqrt k}}{\sqrt n}=\lim_{n\to\infty}\frac 1n\sum_{k=1}^{n}\frac{1}{\sqrt{\frac kn}}=\int_{0}^{1}\frac{1}{\sqrt x}\ dx=2.$$
A: Use the the inequalities:
$$ \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\leq\frac{1}{2\sqrt n} \leq  \frac{1}{\sqrt{n}+\sqrt{n-1}}=\sqrt{n}-\sqrt{n-1}$$
and as a consequence:
$$2\frac{\sqrt{n+1}-1}{\sqrt n}\leq {\sum_{k=1}^n {1\over\sqrt{k}}\over\sqrt{n}}\leq 2$$
and the limit is $2$
A: I'll throw this solution in the mix: Let $S_n$ be the sum in the numerator. Using crayons and rectangles, we see$$\int_1^n \frac{dx}{\sqrt x} < S_n < 1+ \int_1^n \frac{dx}{\sqrt x}.$$The integral equals $2\sqrt n - 2.$ Divide by $\sqrt n$ and the limit of $2$ follow easily.
A: Hint: You can use L'Hopital's rule. So the limit becomes
$$\lim_{n \to \infty} \frac{2 \sqrt{n} }{\sqrt{n}} =2.$$
