# how to prove this series

• May 25th 2008, 06:28 PM
szpengchao
how to prove this series
limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)
• May 25th 2008, 07:23 PM
mr fantastic
Quote:

Originally Posted by szpengchao
limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

How rigorous does the proof need to be?

$\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{n}} = \sum_{k=1}^{\infty} \frac{1}{\sqrt{n}} = + \infty$.
• May 25th 2008, 07:31 PM
Reckoner
Quote:

Originally Posted by szpengchao
limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

What are you trying to prove? I see no propositions in your post. Are you trying to show divergence? We have

$\sum_{k=1}^\infty\sqrt{\frac1k} = \sum_{k=1}^\infty\frac1{\sqrt{k}} = \sum_{k=1}^\infty k^{-1/2}$

This is a p-series, with $p = \frac12$, so you should know that it diverges. If you want to prove it without relying on the properties of p-series, one way is to use the Integral Test. Show that $\int_1^\infty x^{-1/2}\,dx$ diverges (and note that $x^{-1/2}$ is positive, decreasing, and continuous on $[1,\,\infty)$).
• May 25th 2008, 08:03 PM
szpengchao
wrong
sorry, the question is wrong, my question is:

limit( sigma( k^(-1/2),k, 1, n)/sqrt(n),n goes to infinity)

i think this limit is 2
but cant find a way to prove it
• May 25th 2008, 09:49 PM
szpengchao
wat
• May 26th 2008, 12:28 AM
CaptainBlack
Quote:
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^{1/2}}$

RonL
• May 26th 2008, 12:43 AM
szpengchao
wrong
mistake again, there is a sqrt of n there...

this is wat i got when finding integral of 1/sqrt(x)
• May 26th 2008, 12:52 AM
CaptainBlack
Quote:

Originally Posted by szpengchao
mistake again, there is a sqrt of n there...

this is wat i got when finding integral of 1/sqrt(x)

So now we are talking about:

$
\lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{k=1}^n \frac{1}{k^{1/2}}
$

are we?

RonL
• May 26th 2008, 01:01 AM
szpengchao
yes
yes...
• May 26th 2008, 03:10 AM
flyingsquirrel
Hello
Quote:

Originally Posted by CaptainBlack
So now we are talking about:

$
\lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{k=1}^n \frac{1}{k^{1/2}}
$

are we?

RonL

Quote:

Originally Posted by szpengchao
yes...

As $x\mapsto \frac{1}{\sqrt{x}}$ is a decreasing function, you can use $\int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{t}} \leq \int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{k}}\, \leq \int_{k-1}^{k}\frac{\mathrm{d}t}{\sqrt{t}} \Longleftrightarrow
\int_{k}^{k+1}\frac{\mathrm{d}t}{\sqrt{t}} \leq \frac{1}{\sqrt{k}} \leq \int_{k-1}^{k}\frac{\mathrm{d}t}{\sqrt{t}}$
.

Adding these inequalities for $k\in[|2,\,n|]$ gives us :

$\int_{2}^{3}\frac{\mathrm{d}t}{\sqrt{t}}+ \int_{3}^{4}\frac{\mathrm{d}t}{\sqrt{t}}+\ldots+ \int_{n}^{n+1}\frac{\mathrm{d}t}{\sqrt{t}}
\leq \sum_{k=2}^n\frac{1}{\sqrt{k}}
\leq
\int_{1}^{2}\frac{\mathrm{d}t}{\sqrt{t}}+ \int_{2}^{3}\frac{\mathrm{d}t}{\sqrt{t}}+\ldots+ \int_{n-1}^{n}\frac{\mathrm{d}t}{\sqrt{t}}$

$1+\int_{2}^{n+1}\frac{\mathrm{d}t}{\sqrt{t}}
\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}
\leq
1+\int_{1}^{n}\frac{\mathrm{d}t}{\sqrt{t}}$

Can you take it from here ?
• May 26th 2008, 04:17 AM
Krizalid
$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum\limits_{k\ ,=\,1}^{n}{\frac{1}{\sqrt{k}}}=\lim_{n\to\infty}\f rac{1}{n}\sum\limits_{k\,=\,1}^{n}{\sqrt{\frac{n}{ k}}}.$

For each $n,$ it's a Riemann sum for $\int_{0}^{1}{\frac{1}{\sqrt{x}}\,dx}$ and the conclusion follows.
• May 26th 2008, 05:10 AM
szpengchao
follow
i can follow the last step...but how can we get that =2 then?
• May 26th 2008, 05:16 AM
szpengchao
proof
i hav proved that 1/sqrt (x) is integrable. but i m trying to show that integral =2sqrt(a) , so i need to show riemanss sum tends to 2sqrt(a) then..that's what i m troubling with
• May 26th 2008, 10:15 AM
Krizalid
Can't you do $\int_0^1\frac1{\sqrt x}\,dx$?