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Math Help - Using the alternating series test for convergence ?

  1. #1
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    Using the alternating series test for convergence ?

    The alternating series test is given as:

    \sum^{\infty}_{r=1}(-1)^{r-1}b_r , where b_n > 0 , converges if (b_n) is a monotone decreasing sequence with limit zero.


    I have 3 problems,I am asked to investigate these for conditional or absolute convergence, or divergence.

    a) \sum^{\infty}_{r=1}(-1)^r(\sqrt{r(r+1)}-r)

    b) \sum^{\infty}_{r=1}\frac{(-1)^r r^{3/2}}{r^2-1}

    c) \sum^{\infty}_{r=1}\frac{(-1)^r r}{e^r}

    for problem a) :

    I think i should transform (a) into the form \sum^{\infty}_{r=1}(-1)^{r-1}(r-\sqrt{r(r+1)}) and then prove that (r-\sqrt{r(r+1)}) is positive, and a monotone decreasing sequence with limit zero, to show that (a) converges. But how exactly can i show \lim_{n\rightarrow \infty}(n-\sqrt{n(n+1)}) = 0 ? Don't know how to find the limit in this form ! Also,to test for absolute convergence of the series, do i merely test the series but exclude the factor (-1)^r ? i.e to test for absolute convergence do i test the convergence of:

    a) \sum^{\infty}_{r=1}(\sqrt{r(r+1)}-r) ?

    for b) :

    Here r cannot be 1! do i just check from r=2? Getting (b) into the form to use the alternating series test i get
    \sum^{\infty}_{r=1}\frac{(-1)^{r-1}r^{3/2}}{1-r^2} which leaves b_n=\frac{n^{3/2}}{1-n^2} to be proven positive,and decreasing to limit zero. I thought about multiplying both numerator and denominator by n,to divide all terms by a dominant term n^3 ? to get \frac{n^(5/2)}{n-n^3}\div n^3 = \frac{n^{-\frac{1}{2}}}{\frac{1}{n^2}-1}
    This seems to tend to zero as n tends to infinity. But now for absolute convergence, do i test \sum^{\infty}_{r=1}\frac{r^{3/2}}{r^2-1} to see if this converges ?
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  2. #2
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    For #a note that  \displaystyle \sqrt{r(r+1)}-r=\frac{1}{\sqrt{	1+\frac{1}{r}}+1}
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  3. #3
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    It took me a while to see how this helps, but thanks !
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  4. #4
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    Quote Originally Posted by punkstart View Post
    [FONT=Times New Roman]
    But now for absolute convergence, do i test \sum^{\infty}_{r=1}\frac{r^{3/2}}{r^2-1} to see if this converges ?
    For large $$ r the absolute value of the terms behave as r^{-1/2} which means that the series cannot converge absolutely.

    CB
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by punkstart View Post
    c) \sum^{\infty}_{r=1}\frac{(-1)^r r}{e^r}
    for all x>0:\ e^x>\dfrac{x^3}{6}

    CB
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