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Math Help - nth partial sums of series + telescoping series

  1. #1
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    nth partial sums of series + telescoping series

    Alright, so I'm having some conceptual trouble. I'll go ahead and post the homework problems but these are the concept in practice. I don't need all of the homework worked. Just pick and choose and help me see what's going on here. I only need to figure out what the nth partial sum is and how to reliably find it.

    Homework problems, prefer help on one or two of the latter 3 but the first 2 are easier if you don't have the time.

    "Find a formula for the nth partial sum of each series and use it to find the series' sum if the series converges."

    2+ 2/3 + 2/9 + 2/27 + .... 2/3^(n-1) + ...

    (Answer: s_n=((2(1-(1/3)^n)) / (1-(1/3) and it converges at 3.)
    and

    1/(2*3) + 1/(3*4) + 1/(4*5) + ... 1/((n+1)(n+2))

    (Answer is: s_n=1/2 - 1/(n+2) converges at 1/2)

    Then part 2 where I got my butt kicked for not understanding partial sums, in the homework,

    "Find a formula for the nth partial sum of the series and use it to determine if the series converges or diverges. If it converges, find the sum."

    Note: I only need help finding s_n right now!

    Summation from n = 1 to infinity of (1/n - 1/(n+1)
    ( Answer: s_n= 1 - 1/n+1, converges at 1 )

    and

    Summation from n=1 to infinity of ln((n+1)^(1/2)) - ln(n^(1/2))

    ( Answer: s_n= ln((n+1)^(1/2)), diverges )

    and

    Summation from n=1 to infinity of 4/((4n-3)(4n+1))
    ( Answer: 1, no s_n given for answer )


    Thanks guys! Pick and choose! Once I get it nailed down on one, the others will come easily!
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  2. #2
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    The series \displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ... is geometric.

    I will demonstrate how to find the sum of the first \displaystyle n terms of any geometric sequence where \displaystyle a_0 is the first term of the sequence and \displaystyle r is the ratio of the sequence.

    The sum of the first \displaystyle n terms, \displaystyle S_n, is given by

    \displaystyle S_n = a_0 + a_0 r + a_0 r^2 + a_0 r^3 + ... a_0 r^n (1)

    Multiply both sides of the equation by \displaystyle r.

    \displaystyle r S_n = a_0 r + a_0 r^2 + a_0 r^3 + a_0 r^4 + ... a_0 r^{n+1} (2)

    Subtract equation 2 from equation 1.

    \displaystyle S_n - r S_n = a_0 - a_0 r^{n + 1} \iff S_n\left(1 - r\right) = a_0\left(1 - r^{n + 1}\right)

    Divide both sides of the equation by \displaystyle 1 - r.

    S_n = \dfrac{a_0 \left(1 - r^{n + 1}\right)}{1 - r}

    Now, it's a matter of recognizing geometric series and identifying \displaystyle a_0 and \displaystyle r. What is \displaystyle a_0 and \displaystyle r for the geometric series \displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ... ?
    Last edited by NOX Andrew; April 12th 2011 at 11:41 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    you mean "converges to 3"

    I think what you need to understand is what a partial sum is.
    It is the sum of the first N terms.

    S_N=a_1+\cdots +a_N

    AND its LIMIT is the infinite sum

    \sum_{i=1}^{\infty}a_i=\lim_{N\to\infty}\sum_{i=1}  ^{N}a_i =\lim_{N\to\infty}S_N
    Last edited by matheagle; April 12th 2011 at 10:05 PM.
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  4. #4
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    The series \displaystyle \sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right) is a telescoping series.

    The sum of the first \displaystyle N terms of a telescopic sequence, \displaystyle S_N, is given by

    \displaystyle S_N = \left(\dfrac{1}{1} - \dfrac{1}{1 + 1}\right) + \left(\dfrac{1}{2} - \dfrac{1}{2 + 1}\right) + \left(\dfrac{1}{3} - \dfrac{1}{3 + 1}\right) + \left(\dfrac{1}{4} - \dfrac{1}{4 + 1}\right) + ...  + \left(\dfrac{1}{N} - \dfrac{1}{N + 1}\right)

    Simplify the right-hand side of the equation.

    \displaystyle S_N = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + ...  + \left(\dfrac{1}{N} - \dfrac{1}{N + 1}\right)

    Apply the associative property of addition to conveniently rearrange the terms.

    \displaystyle S_N = 1 + \left(-\dfrac{1}{2} + \dfrac{1}{2}\right) + \left(-\dfrac{1}{3} + \dfrac{1}{3}\right) + \left(-\dfrac{1}{4} + \dfrac{1}{4}\right) + ... + \left(-\dfrac{1}{N} + \dfrac{1}{N}\right) - \dfrac{1}{N + 1}

    All but the first and last terms cancel.

    \displaystyle S_N = 1 - \dfrac{1}{N + 1}

    A general telescopic series looks like \displaystyle \sum\limits_{n = 1}^{\infty} \left(a_n - a_{n + 1}\right). The sum of the first \displaystyle N terms of a telescopic series, \displaystyle S_N, is given by

    \displaystyle S_N = a_1 - a_{N + 1}

    In words, the sum of the first \displaystyle N terms of a telescopic sequence is the sum of the first term and the \displaystyle (N + 1)th term (because all the terms in between cancel).

    Like before, it's now a matter of recognizing telescopic series and identifying \displaystyle a_1 and \displaystyle a_{N + 1}.

    The remaining series are telescopic - algebraic manipulation helps to recognize them as much. Don't hesitate to ask any more questions. However, I suggest reviewing the section on series in your textbook as all of your problems could have been solved with knowledge of common series formulas.
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  5. #5
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    Didn't see your post, Andrew, I'd left it up overnight! Will double reply soon.

    Original post:
    Oops @ the geometric series. We were not shown that s_n nor did he work a problem like this. He pretty much dropped summation n to infinity of a * r^n-1 = a/1-r if |r|<1 and moved on.

    So what is the quick and dirty way to always find the nth partial sum so that I can take the limit to see where a series converges? Both when given a set of numbers and when given an actual series?

    It's likely he's going to quiz tonight regardless of how many people in the class are baffled, so I need to get this patched up and figured out.
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