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Math Help - Converge / Diverge

  1. #1
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    Converge / Diverge

    Which ones are convergent and which are divergent and what are the limits of the convergent ones?
    a) lim (as n goes to infinity) of (-1)^n sin(n)/n

    b) lim (as n goes to infinity) of (-1)^n sin(1/n)

    c) lim (as n goes to infinity) of n^(1/n) (L'Hospital's Rule)

    d) lim (as n goes to infinity) a(sub)n where a1 = sqrt(2) , a2 = sqrt(2sqrt(2)), a3 = sqrt(2sqrt(2sqrt2))) , an+1 = sqrt(2an)

    i think a is divergent becaue of the alternator and b converges to 0 because of squeeze theorem. Is this right? And how do i find the other two?
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  2. #2
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    For a, look at the \displaystyle\frac{sin(n)}{n}. What happens when n goes to infinity there?

    For b, concentrate on the sin portion.

    For c, set equal to y and natural log.

    For d, I think this should work \displaystyle x=\sqrt{2\sqrt{2\sqrt{2....}}}
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  3. #3
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    so the first two converge to 0?
    lny = (1/n) ln (n)? it says to use L'Hospital's Rule for c
    im not sure what you mean for d, but would it just diverge?
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  4. #4
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    First 2 are 0.

    Ok so now we have, for the 3rd, \displaystyle \lim_{n\to\infty}\frac{ln(n)}{n}=\frac{\infty}{\in  fty} So now use L'Hopital's Rule. Remember to exponentiate the answer to undo the natural log.

    However, you should know this goes to zero since y=x grows faster than y=ln(x).

    \displaystyle x=\sqrt{2\sqrt{2\sqrt{2....}}}\rightarrow x^2=2\sqrt{2\sqrt{2\sqrt{2....}}} \ \mbox{but x=\sqrt{2\sqrt{2\sqrt{2....}}}} \ x^2=2x
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Playthious View Post
    Which ones are convergent and which are divergent and what are the limits of the convergent ones?
    a) lim (as n goes to infinity) of (-1)^n sin(n)/n

    b) lim (as n goes to infinity) of (-1)^n sin(1/n)

    c) lim (as n goes to infinity) of n^(1/n) (L'Hospital's Rule)

    d) lim (as n goes to infinity) a(sub)n where a1 = sqrt(2) , a2 = sqrt(2sqrt(2)), a3 = sqrt(2sqrt(2sqrt2))) , an+1 = sqrt(2an)

    i think a is divergent becaue of the alternator and b converges to 0 because of squeeze theorem. Is this right? And how do i find the other two?
    The solution of d) is fully elementar ifn someone follows the procedure I illustrated a lot of times. The 'recursive' equation can be written as...

    \displaystyle \Delta_{n} = a_{n-1}-a_{n} = \sqrt{2 a_{n}} - a_{n} = f(a_{n}) (1)

    The function f(x)= \sqrt{2 x} - x is represented here...



    There is only one 'attractive fixed point' in x_{0}=2 and because is |f(x)|< |x_{0}-x| [see 'red line'...] any a_{0}>0 will generated a sequence monotonically convergent at x_{0}...

    Kind regards

    \chi \sigma
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