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Math Help - sequence- diverges/converges and limits

  1. #1
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    sequence- diverges/converges and limits

    Hello
    I am trying to figure out if the following sequence converges:
    Pn={(1/n)*sin(npi/2))} from n=1 to infinity.

    This is how I am approaching it. I am using the fact that a sequence converges iff all of its subsequences converge to the same limit. So, I assume that Pn converges to p.
    Then, I look at the subsequence {sin(npi/2)}. This subsequence doesn't converge because it fluctuates between -1,0 and 1 depending on the value of n. So, since this sequence doesn't have a limit and it doesn't converge, then Pn diverges. Am I right?

    Thanks
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  2. #2
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    Quote Originally Posted by inthequestofproofs View Post
    Hello
    I am trying to figure out if the following sequence converges:
    Pn={(1/n)*sin(npi/2))} from n=1 to infinity.

    This is how I am approaching it. I am using the fact that a sequence converges iff all of its subsequences converge to the same limit. So, I assume that Pn converges to p.
    Then, I look at the subsequence {sin(npi/2)}. This subsequence doesn't converge because it fluctuates between -1,0 and 1 depending on the value of n. So, since this sequence doesn't have a limit and it doesn't converge, then Pn diverges. Am I right?

    Thanks
    You need to be careful what happend to the factor \frac{1}{n}.

    So you have the product of two sequences one of which is bounded and the other is going to zero...
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  3. #3
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    so, if the lim 1/n = 0 and the lim sin(npi/2) doesn't exist, then can i use the fact that lim(1/n)(sin(npi/2)) = lim1/n * lim sin(npi/2)? or is this property only used for sequences that converge?

    how do i go about proving if the sequence converges?
    SHould I find a p and an n0 such that |pn - p| < e for all n>= n0?

    Thanks!
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by inthequestofproofs View Post
    so, if the lim 1/n = 0 and the lim sin(npi/2) doesn't exist, then can i use the fact that lim(1/n)(sin(npi/2)) = lim1/n * lim sin(npi/2)? or is this property only used for sequences that converge?

    how do i go about proving if the sequence converges?
    SHould I find a p and an n0 such that |pn - p| < e for all n>= n0?

    Thanks!
    You should have a theorem about the first hint that I gave you.
    As an alternative you could use the squeeze theorem.

    \frac{-1}{n}\le P_n \le \frac{1}{n}

    Or just write out some terms and see if you can find a pattern. This have an equivilent form that is much simpler.

    Good luck.
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  5. #5
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    I got it!
    using the squeeze theorem, i know lim -1/n = 0 and lim1/n =0 thus, the lim pn =0 by squeeze theorem.

    Thanks
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  6. #6
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    i have another question for you:
    how do i show that pn={(ncos(npi))/(2n+3)} converges or diverges.

    THis is what i did so far:
    [ncos(npi)][/(2n+3)]=[cos(npi)][/(2+(3/n)]

    as n --> infinity, the numerator fluctuates from -1 and 1. The denominator, goes to 2. So, as n goes to infinity, the sequences takes the values -1/2 or 1/2 depending on the cos(npi).

    Does this sound good to prove that it diverges?

    THanks
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  7. #7
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    Why make so difficult?
    \frac{{ - 1}}<br />
{n} \leqslant \frac{1}<br />
{n}\sin \left( {\frac{{n\pi }}<br />
{2}} \right) \leqslant \frac{1}<br />
{n}
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