can't determine the convergence/divergence here

Question

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 $r$

The sum of the reciprocals of the square roots is about $2\sqrt{n}$. Divide by $n$. That gives $t_n$ about $2/\sqrt{n}$, so $\sum t_n$ diverges. — André Nicolas, Feb 04 '16 at 07:31
Actually, $t_n$ seems to converge with a little bit of testing — , Feb 04 '16 at 07:33

Mark Schultz-Wu · Answer 1 · 2016-02-04T07:37:30.947

2

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.

edited Feb 04 '16 at 07:37

answered Feb 04 '16 at 07:33

Mark Schultz-Wu

7,234

$t_{100000}=0.0063099675866237874808483$, not very convincing divergence – Feb 04 '16 at 07:35
But the argument of Mark is fully convincing. – André Nicolas Feb 04 '16 at 07:37
I feel like I miss the factor $\frac 1n$ before the sum – Feb 04 '16 at 07:37
I've edited to include that I was referencing only the sum. $t_n$ itself may still converge, and it appears there's an answer that shows that currently. – Mark Schultz-Wu Feb 04 '16 at 07:39
Right, Mark, excuse me. I was completely missing the point you were trying to get across. My mistake! – Feb 04 '16 at 07:40
@Mark: $t_n$ converges to $0$, but the question was about $\sum t_n$. – André Nicolas Feb 04 '16 at 07:41

Ángel Mario Gallegos · Answer 2 · 2016-02-04T14:49:09.460

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.

Martin Argerami · Accepted Answer · 2016-02-04T17:27:16.703

2

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:$$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\left (\frac1n\,\sum_{k=1}^n\frac1 {\sqrt{ k/n}}\right)\,\lim_{n\to\infty}\frac1 {\sqrt n}=2\times0=0. $$

edited Feb 04 '16 at 17:27

answered Feb 04 '16 at 07:45

Martin Argerami

217,281

so does the sequence converges.thanks for help – Onix Feb 04 '16 at 08:06
@Abomm Another way to see that $t_n$ converges to zero is to notice that $1/\sqrt n\to 0$. Then the same is true for the averages: http://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means – Martin Sleziak Feb 04 '16 at 15:30

score 0 · Answer 4 · answered Feb 04 '16 at 08:06

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.

can't determine the convergence/divergence here

4 Answers4