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 $r$
 A: You have $\sum_{n=1}^{\infty}t_n$.  Note that $t_n-\frac{1}{n}\geq 0$, with equality only at $n=1$.  So, you have a series that's greater than or equal to a divergent series (the harmonic series), so by the comparison test it (the sum) is divergent.
A: If $n\ge1$ is an integer then $n\sqrt{k}\ge k\sqrt{k}$ for $1\le k\le n$, then
$$t_n=\sum_{k=1}^{n}\frac{1}{n\sqrt{k}}\le\sum_{k=1}^n\frac{1}{k^{3/2}}$$
Last one is the $p$-series, where $p=3/2$, so $t_n$ converges as $n\to\infty$.
On the other hand, $t_1=1$ and $t_n>1/n$ for $n>1$, then $$\sum_{n=1}^{M}t_n>\sum_{n=1}^{M}\frac{1}{n}\qquad\text{for all integer }M>1$$
Since the harmonic series diverges it follows, from the comparison test, that also $\sum t_n$ diverges.
A: 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:$$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\left (\frac1n\,\sum_{k=1}^n\frac1 {\sqrt{ k/n}}\right)\,\lim_{n\to\infty}\frac1 {\sqrt n}=2\times0=0. $$
A: Using Abel's summation we can easily see that $$\sum_{k\leq n}\frac{1}{\sqrt{k}}=\sum_{k\leq n}1\cdot\frac{1}{\sqrt{k}}=\sqrt{n}+\frac{1}{2}\int_{1}^{n}\left\lfloor t\right\rfloor t^{-3/2}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Then, using the estimation $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we have $$\sum_{k\leq n}\frac{1}{\sqrt{k}}=2\sqrt{n}+O\left(\frac{1}{\sqrt{n}}\right)
 $$ and now I think you can conclude by yourself.
