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Math Help - Convergent/Divergent Sequences...

  1. #1
    zee
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    Convergent/Divergent Sequences...

    Which of the following sequences are convergent and which are divergent?
    a) a_{n} = (1+2n)^{\frac{1}{n}}
    b) a_{n} = cos(n\pi)
    c) a_{n} = \frac{cos(3n)}{(1+n^{1/2})}

    This is what I got:
    a) By the squeeze theorem, 1 < (1+2n)^1/2 < 3
    and the sequence decreases as a(subscript n) > a(subscript n+1)
    So is it correct to say that the sequence diverges?

    b) {a(subscript n)} = {cosPI, cos2PI, cos3PI, cos4PI...}
    = {-1,1,-1,1,...}
    Since this sequence oscillates between -1 and 1, so it diverges.
    Is this correct?

    c) since cos(3n) <= 1 for all n, we have
    cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
    and we know that 1/n^1/2 is divergent (since p=1/2)
    Is this correct?

    I am almost sure that b and c are correct, but as for question a,
    im lost... please help.. zee
    Last edited by MathGuru; May 28th 2005 at 07:53 AM. Reason: latex
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  2. #2
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    (a) Diverge. The numerator is always larger than the denominator.
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  3. #3
    hpe
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    Quote Originally Posted by zee
    Which of the following sequences are convergent and which are divergent?
    a) a(subscript n) = (1+2n)^1/n
    b) a(subscript n) = cos(nPI)
    c) a(subscript n) = cos(3n)/(1+n^1/2)

    This is what I got:
    a) By the squeeze theorem, 1 < (1+2n)^1/2 < 3
    and the sequence decreases as a(subscript n) > a(subscript n+1)
    You need to prove that.
    So is it correct to say that the sequence diverges?
    The correct conclusion is that the sequence converges .

    b) {a(subscript n)} = {cosPI, cos2PI, cos3PI, cos4PI...}
    = {-1,1,-1,1,...}
    Since this sequence oscillates between -1 and 1, so it diverges.
    Is this correct?
    Yes. But check if you need a more formal argument.
    c) since cos(3n) <= 1 for all n, we have
    cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
    and we know that 1/n^1/2 is divergent (since p=1/2)
    Is this correct?
    Incorrect. This is a sequence, not a series. Use the squeeze theorem.

    Btw, would you please confirm here in writing that these are not questions from a take-home final?
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  4. #4
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    If the OP meant (1+2n)^{(1/n)} then yes the sequence converges.
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  5. #5
    zee
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    more confused...


    I dont understand why (1+2n)^1/n is convergent.
    I did apply the squeeze theorem as above
    1 < (1+2n)^1/2 < 3
    but still it turns out to be divergent because the
    series is decreasing. I even plotted a table to check
    if it was decreasing. What am I doing wrong?

    Also for the other question, cos(3n)/(1+n^1/2):
    I applied the squeeze theorem here too.
    cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
    as above. And the rule is:
    For: 1/n^p, if p>1, it is convergent, if p<=1, it's divergent.
    In this question it is clearly divergent as p = 1/2.

    I know i must be dong something wrong as I'm getting
    the answers for all the practice questions to be divergent.
    I do have all the criterias for when they are div and conv,
    but somehow the end answers always satisfy the div property.
    Please assist... becomin more confused!
    zee
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  6. #6
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    Please write as (1+2n)^(1/2) if you meant it to be a square root. This might also be your source of error.
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  7. #7
    zee
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    yes...

    yes that is correct...
    i did mean (1+2n)^(1/2) ....
    though workin it by hand.. there were no errors...
    any suggestionss??
    zee
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  8. #8
    hpe
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    Also for the other question, cos(3n)/(1+n^1/2):
    I applied the squeeze theorem here too.
    cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
    as above. And the rule is:
    For: 1/n^p, if p>1, it is convergent, if p<=1, it's divergent.
    In this question it is clearly divergent as p = 1/2.
    Your estimate is correct:
    <br />
\left|\frac{\cos(3n)}{1+\sqrt{n}}\right| \le \frac{1}{1+\sqrt{n}} \le \frac{1}{\sqrt{n}}<br />
    Therefore
    <br />
\lim_{n \to \infty} \frac{\cos(3n)}{1+\sqrt{n}} = 0<br />
    by the squeeze theorem. This is a limit of a sequence.

    However
     \sum_{n=1}^\infty \frac{1}{\sqrt{n}}<br />
    does not converge (p-series). This is an infinite series .

    Check what your original question was. I suspect it was about limits of a sequence. So you shouldn't be thinking in terms of p-series in this context.
    I guess that is why you are confused.

    This does not, however, settle whether  \sum_{n=1}^\infty \frac{\cos(3n)}{1+\sqrt{n}} converges or diverges. However, it cannot converge absolutely. I suspect it converges conditionally - anybody ready to prove this?
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