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Math Help - convergent sequences

  1. #1
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    convergent sequences

    Hi,

    Can someone explain to me why [sin(0.5*npi)]/n converges to 0.

    The explanation I have been given is that since it is < 1/n and therefore can be made as small as we like for large enough n.

    But I don't follow why it is smaller than 1/n. I guess if I understood this it would make sense that as n tends to infinity, the fn. tends to 0.


    I am also confused from reading wolfram articles how to evaluate a geometric series and how to tell whether one converges or not. Could someone maybe explain to me how to evalaute:

    a) sum between 2&9 of 10^-n
    &
    b) sum between 0&ininfity of 3^(-n/2)
    &
    c) determine whether convergent or not:
    Between 1&infinity [(n-1)/n]

    many thanks
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  2. #2
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    Quote Originally Posted by Bryn View Post
    Hi,

    Can someone explain to me why [sin(0.5*npi)]/n converges to 0.

    The explanation I have been given is that since it is < 1/n and therefore can be made as small as we like for large enough n.

    But I don't follow why it is smaller than 1/n.
    Actually, it isn't smaller- it is "less than or equal to". That is, sin(x)/n\le 1/n, for any x whatsoever, because -1\le sin(x)\le 1 for all x.

    I guess if I understood this it would make sense that as n tends to infinity, the fn. tends to 0.


    I am also confused from reading wolfram articles how to evaluate a geometric series and how to tell whether one converges or not. Could someone maybe explain to me how to evalaute:

    a) sum between 2&9 of 10^-n
    &
    b) sum between 0&ininfity of 3^(-n/2)
    &
    Those are both "geometric" sequences:
    \sum_{i=0}^n r^i= \frac{1- r^{n-1}}{1- r}
    And, of course, \sum_{i= 2}^9= \sum_{i=0}^9- \sum_{i=0}^2.

    c) determine whether convergent or not:
    Between 1&infinity [(n-1)/n]
    Do you mean the sum? it's fairly easy to see that \frac{n-1}{n}= \frac{n}{n}- \frac{1}{n}= 1- \frac{1}{n} and that sequence converges to 1 as n goes to infinity.

    Do you know this theorem: If \sum{n=0}^\infty a_n converges, then \lim_{n\to\infty} a_n= 0?
    If you you don't recognise that, realize that after n= 100000000000, say, you are still adding numbers close to 1.

    many thanks
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  3. #3
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    Hello Bryn
    Quote Originally Posted by Bryn View Post
    Hi,

    Can someone explain to me why [sin(0.5*npi)]/n converges to 0.

    The explanation I have been given is that since it is < 1/n and therefore can be made as small as we like for large enough n.

    But I don't follow why it is smaller than 1/n. I guess if I understood this it would make sense that as n tends to infinity, the fn. tends to 0.
    You are right to question this: it doesn't converge to 0; \lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}

    It is well-known that \lim_{x\to 0}\;\frac{\sin(x)}{x}=1, so if we replace x by (n\pi/2), \frac{\sin(x)}{x}=\frac{\sin(n\pi/2)}{(n\pi/2)}= \frac{2}{\pi}\cdot\frac{\sin(n\pi/2)}{n}, and we note that as x \to 0, n\to0.

    Therefore
    1=\lim_{x\to 0}\;\frac{\sin(x)}{x}
    =\lim_{n\to0}\frac{2}{\pi}\cdot\frac{\sin(n\pi/2)}{n}
    =\frac{2}{\pi}\cdot\lim_{n\to0}\frac{\sin(n\pi/2)}{n}
    \Rightarrow\lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}
    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello Bryn[SIZE=3]You are right to question this: it doesn't converge to 0; \lim_{n\to 0}\;\frac{\sin(n\pi/2)}{n}=\frac{\pi}{2}
    It should be <br />
\lim_{n\to \color{blue}\infty}\;\frac{\sin(n\pi/2)}{n}
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  5. #5
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    Quote Originally Posted by Plato View Post
    It should be <br />
\lim_{n\to \color{blue}\infty}\;\frac{\sin(n\pi/2)}{n}
    Ah yes, of course, you're right! I suppose I thought it was n\to 0 because it's not so obvious what the limit is in that case.

    I should have read it a bit more carefully!

    Grandad
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