<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] Note that your sequence can be written as follows
$$t_n = \frac{1}{n}\sum_{j=1}^n\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/2}}dt\; + \frac{1}{\sqrt{n+1}}.$$
Clearly the last term of the summation vanishes as $n\to+\infty$ while the first term is the Cesaro sum of the sequence
$$a_j:=\frac{j}{2}\int_{j}^{j+1}\frac{1}{t^{3/2}}dt.$$
By a direct computation you can verify that $\lim_{j\to+\infty}a_j=0$ and by Cesaro theorem this means that $1/n\sum_{j=1}^n a_j$ converges to the same limit as $n\to+\infty$ and this shows that $t_n\to 0$.
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