<s>[INST] Calculate the limit (Squeeze Theorem?) 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.
 [/INST] 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
$$ 2 \sqrt {2n+1} - 2 \sqrt {n+1} < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2n} - 2 \sqrt {n}  $$
$$ 2 \sqrt {2+\frac{1}{n}} - 2 \sqrt {1+\frac{1}{n}} < \frac{1}{\sqrt n} \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2} - 2 \sqrt {1}  $$
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