<s>[INST] Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ using AM-GM inequality, }\space\space t_n\ge (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\le t_n\le \dfrac{1}{n}(1+\cdots+1)=\dfrac{n}{n}=1$
$(1\cdot2\cdot 3\cdots n)^{\dfrac{1}{n}}=(1)^{\frac{1}{n}}\cdot (2)^{\frac{1}{n}}\cdot (3)^{\frac{1}{n}}\cdots\cdot (n)^{\frac{1}{n}}\stackrel{n\to\infty}{\to}1$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\to1$
Hence, by sandwich theorem $t_n\to1$
Is this correct?
 [/INST] There is a Riemann sum in disguise.
Note, however, that the function to integrate is unbounded. It still works with the improper integral, as is explained here: Convergence of Riemann sums for improper integrals.
$$t_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt k}=\frac{1}{n}\sum_{k=1}^n\frac{1/\sqrt{n}}{\sqrt {k/n}}=\frac{1}{n\sqrt n}\sum_{k=1}^n\frac{1}{\sqrt {k/n}}$$
Hence,
$$\sqrt{n}\,t_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt {k/n}}\to\int_0^1\frac{1}{\sqrt x}\mathrm dx=2$$
And
$$t_n\sim \frac{2}{\sqrt n}$$
</s>