limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

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- May 25th 2008, 06:28 PMszpengchaohow to prove this series
limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)

- May 25th 2008, 07:23 PMmr fantastic
- May 25th 2008, 07:31 PMReckoner
What are you trying to prove? I see no propositions in your post. Are you trying to show divergence? We have

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

This is a*p*-series, with $\displaystyle p = \frac12$, 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 $\displaystyle \int_1^\infty x^{-1/2}\,dx$ diverges (and note that $\displaystyle x^{-1/2}$ is positive, decreasing, and continuous on $\displaystyle [1,\,\infty)$). - May 25th 2008, 08:03 PMszpengchaowrong
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 PMszpengchaowat
- May 26th 2008, 12:28 AMCaptainBlack
- May 26th 2008, 12:43 AMszpengchaowrong
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 AMCaptainBlack
- May 26th 2008, 01:01 AMszpengchaoyes
yes...

- May 26th 2008, 03:10 AMflyingsquirrel
Hello

As $\displaystyle x\mapsto \frac{1}{\sqrt{x}}$ is a decreasing function, you can use $\displaystyle \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 $\displaystyle k\in[|2,\,n|]$ gives us :

$\displaystyle \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}}$

Adding 1 yields

$\displaystyle 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 AMKrizalid
$\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum\limits_{k\ ,=\,1}^{n}{\frac{1}{\sqrt{k}}}=\lim_{n\to\infty}\f rac{1}{n}\sum\limits_{k\,=\,1}^{n}{\sqrt{\frac{n}{ k}}}.$

For each $\displaystyle n,$ it's a Riemann sum for $\displaystyle \int_{0}^{1}{\frac{1}{\sqrt{x}}\,dx}$ and the conclusion follows. - May 26th 2008, 05:10 AMszpengchaofollow
i can follow the last step...but how can we get that =2 then?

- May 26th 2008, 05:16 AMszpengchaoproof
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 AMKrizalid
Can't you do $\displaystyle \int_0^1\frac1{\sqrt x}\,dx$?