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Math Help - how to prove this series

  1. #1
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    how to prove this series

    limit( Sigma( sqrt(1/k),k goes from 1 to n), n goes to infinity)
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    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.
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  3. #3
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    Quote Originally Posted by szpengchao View Post
    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)).
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  4. #4
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    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
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  5. #5
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    wat

    Attached Files Attached Files
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  6. #6
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    Quote Originally Posted by szpengchao View Post
    \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^{1/2}}

    RonL
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  7. #7
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    wrong

    mistake again, there is a sqrt of n there...

    this is wat i got when finding integral of 1/sqrt(x)
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  8. #8
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    Quote Originally Posted by szpengchao View Post
    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:

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

    are we?

    RonL
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  9. #9
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    yes

    yes...
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  10. #10
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    Hello
    Quote Originally Posted by CaptainBlack View Post
    So now we are talking about:

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

    are we?

    RonL
    Quote Originally Posted by szpengchao View Post
    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 <br />
\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}}<br />
\leq \sum_{k=2}^n\frac{1}{\sqrt{k}}<br />
\leq <br />
\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

     1+\int_{2}^{n+1}\frac{\mathrm{d}t}{\sqrt{t}}<br />
\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}<br />
\leq <br />
1+\int_{1}^{n}\frac{\mathrm{d}t}{\sqrt{t}}

    Can you take it from here ?
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  11. #11
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    \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.
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  12. #12
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    follow

    i can follow the last step...but how can we get that =2 then?
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  13. #13
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    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
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  14. #14
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    Can't you do \int_0^1\frac1{\sqrt x}\,dx?
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