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?
 A: You can calculate the limit in an elementary way using
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt k}
& = & \frac 1n\left(1+ \sum_{k=1}^{n-1} \frac 1{\sqrt{k+1}}\right) \\
& < & \frac 1n\left(1+ 2\sum_{k=1}^{n-1} (\sqrt{k+1}-\sqrt k)\right) \\
& = & \frac 1n\left(1+ 2(\sqrt{n}-1)\right) \\
& = & -\frac 1n + \frac 2{\sqrt n} \stackrel{n\to\infty}{\longrightarrow} 0
\end{eqnarray*}
A: The problem with your solution is: $(1.2.3...n)^{\frac 1n}\to 1$. Note that this is false.
Infact, $(1.2.3...n)^{\frac 1n}=n!^{\frac 1n}\sim \frac n e$ by Stirling's approximation.
To evaluate the limit in OP, you may use this application of Stolz-Cesàro's theorem.
A: 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
$$t_n\sim \frac{2}{\sqrt n}$$
A: In a class of $n$ students the $j$th student had a grade of $g_j$ for each $1\le j\le n,$ and the class-average grade was $t_n.$ An $(n+1)$th student was added, and his grade $g_{n+1}$ was less than any of the others' grades. So the class average $t_{n+1}$ went down.
In other words, let $S_n=\sum_{j=1}^n g_j$ and $S_{n+1}=\sum_{j=1}^{n+1} g_j$. We have $$t_{n+1}<t_n \iff S_{n+1}/(n+1)<S_n/n\iff$$ $$\iff nS_{n+1}<(n+1)S_n\iff$$ $$\iff n(g_{n+1}+S_n)<(n+1)S_n\iff$$ $$\iff ng_{n+1}<S_n$$ and the last line above is true because $$S_n=\sum_{j=1}^n g_j\ge\sum_{j=1}^n \min_{j\le n} g_j=$$ $$=n\cdot \min_{j\le n}g_j>n\cdot g_{n+1}.$$
In your Q we have $g_j= 1/\sqrt j$. So $0<t_{n+1}<t_n$ for each $n$. A decreasing positive sequence must converge.
Your attempt to show that $(n!)^{1/n}\to 1$ is wrong. E.g. if $n=2m $ is even then $n!^{1/n}=((2m)!)^{1/2m}\ge ((m+1)...(2m))^{1/2m}\ge ((m+1)^m)^{1/2m}=\sqrt {m+1}.$ E.g. $(1000000!)^{1/1000000}\ge\sqrt {500001}.$
Fallacy of the interchange of order of limits: For $i,j\in \Bbb N$ let $g_{i,j}=j^{1/i}$ when $j\le i$ and let $g_{i,j}=1$ when $j>i.$ Let $P(n,m)=\prod_{j=1}^mg_{n,j}.$ We have $$\lim_{m\to\infty} \lim_{n\to\infty}P(n,m)=1$$  but we do NOT have $\lim_{n\to\infty}\lim_{m\to\infty} P(n,m)=1$ because $\lim_{m\to\infty}P(n,m)=(n!)^{1/n}.$
A: 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 $t_n\to 0$.
