I have to calculate the limit of this formula as $n\to \infty$.
$$a_n = \frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$\frac{1}{\sqrt{2}}\leftarrow\frac{n}{\sqrt{2n^2}}\le\frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl) \le \frac{n}{\sqrt{n^2+n}}\to1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
Rearrange it as $$ \frac{1}{n}\left(\sqrt{\frac{n}{n+1}}+\sqrt{\frac{n}{n+2}}+\ldots+\sqrt{\frac{n}{n+n}}\right) = \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}$$ which is a Riemann sum for $$ \int_{0}^{1}\frac{dx}{\sqrt{1+x}}=2\sqrt{2}-2.$$ Since $\frac{1}{\sqrt{1+x}}$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that $\{a_n\}_{n\geq 1}$ is an increasing sequence convergent to $2\sqrt{2}-2$. Additionally it is not difficult to check that $a_n= 2\sqrt{2}-2-\Theta\left(\frac{1}{n}\right)$ as $n\to +\infty$.
As an alternative by Stolz-Cesaro
$$\frac{b_n}{c_n} = \frac{\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}}{\sqrt n}$$
$$\frac{b_{n+1}-b_n}{c_{n+1}-c_n} = \frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt n}$$
and
$$\frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt n}\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}=$$
$$\frac{\sqrt{n+1}+\sqrt n}{\sqrt{2n+2}}+\frac{\sqrt{n+1}+\sqrt n}{\sqrt{2n+1}}-\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}}\to\frac4{\sqrt 2}-2=2\sqrt 2-2$$
for a decreasing function such as $1/\sqrt x$ with $x$ positive, a simple picture shows $$ \int_a^{b+1} \; f(x) \; dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b} \; f(x) \; dx $$ $$ \int_{n+1}^{2n+1} \; \frac{1}{\sqrt x} \; dx < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < \int_{n}^{2n} \; \frac{1}{\sqrt x} \; dx $$ getting there $$ 2 \sqrt {2n+1} - 2 \sqrt {n+1} < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2n} - 2 \sqrt {n} $$ $$ 2 \sqrt {2+\frac{1}{n}} - 2 \sqrt {1+\frac{1}{n}} < \frac{1}{\sqrt n} \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2} - 2 \sqrt {1} $$
