Wsrtjtyk

I have to calculate the limit of this formula as $nto infty$.

$$a_n = frac{1}{sqrt{n}}bigl(frac{1}{sqrt{n+1}}+cdots+frac{1}{sqrt{2n}}bigl)$$

I tried the Squeeze Theorem, but I get something like this:

$$frac{1}{sqrt{2}}leftarrowfrac{n}{sqrt{2n^2}}lefrac{1}{sqrt{n}}bigl(frac{1}{sqrt{n+1}}+cdots+frac{1}{sqrt{2n}}bigl) le frac{n}{sqrt{n^2+n}}to1$$

As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.

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

$$frac{frac{1}{sqrt{2n+2}}+frac{1}{sqrt{2n+1}}-frac{1}{sqrt{n+1}}}{sqrt{n+1}-sqrt n}frac{sqrt{n+1}+sqrt n}{sqrt{n+1}+sqrt n}=$$

$$frac{sqrt{n+1}+sqrt n}{sqrt{2n+2}}+frac{sqrt{n+1}+sqrt n}{sqrt{2n+1}}-frac{sqrt{n+1}+sqrt n}{sqrt{n+1}}tofrac4{sqrt 2}-2=2sqrt 2-2$$

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
$$ int_{0}^{1}frac{dx}{sqrt{1+x}}=2sqrt{2}-2.$$
Since $frac{1}{sqrt{1+x}}$ is a convex function on $[0,1]$, the Hermite-Hadamard and Karamata's inequalities give us that ${a_n}_{ngeq 1}$ is an increasing sequence convergent to $2sqrt{2}-2$. Additionally it is not difficult to check that $a_n= 2sqrt{2}-2-Thetaleft(frac{1}{n}right)$ as $nto +infty$.

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
$$ 2 sqrt {2n+1} - 2 sqrt {n+1} < sum_{k=n+1}^{2n} frac{1}{sqrt k} < 2 sqrt {2n} - 2 sqrt {n} $$
$$ 2 sqrt {2+frac{1}{n}} - 2 sqrt {1+frac{1}{n}} < frac{1}{sqrt n} sum_{k=n+1}^{2n} frac{1}{sqrt k} < 2 sqrt {2} - 2 sqrt {1} $$