sequences and series - can't determine the convergence/divergence here

Let $$t_{n}=\frac{1}{n}\left(1+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}\right),\ n=1,2,\dots$$ then I want to know if $\sum_{n=1}^{\infty}t_{n}$ converges/diverges and the sequence$\{t_{n}\}$ converges and diverges
for it I thought of finding $\lim_{n\to\infty}t_{n}\\=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\frac{1}{\sqrt{r}}$ but how to solve this limit I can do it if it is presented as a Riemann sum like if there is $n$ in the denominator of $r$

Answer

First $t_n>1/n $, so $\sum_{n=1}^\infty=\infty $.

For $t_n $ alone,

$$
t_n
=\frac1n\,\sum_{k=1}^n\frac1 {\sqrt k}
=\left (\frac1n\,\sum_{k=1}^n\frac1 {\sqrt{ k/n}}\right)\,\frac1 {\sqrt n}.
$$
The expression in brackets converges to $\int_0^1\frac1 {\sqrt t}\,dt=2 $, so the product converges to zero:$$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\left (\frac1n\,\sum_{k=1}^n\frac1 {\sqrt{ k/n}}\right)\,\lim_{n\to\infty}\frac1 {\sqrt n}=2\times0=0. $$