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Math Help - infinite series

  1. #1
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    infinite series

    what's the difference between these two series?

    1.  1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-...


    2. (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})+...

    i think i understand how the 2nd one goes , but i don't get the first one. what difference does that last term make? and like in the 2nd one, the  (\frac{1}{n}-\frac{1}{n+1})towards the end gives the general term. but \frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1} doesn't seem like the general term for the 1st series.
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  2. #2
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    Re: infinite series

    The "general term" is a_n= \frac{1}{n/2}= \frac{2}{n} when when n is even, and (-1)\frac{2}{n-1} when n is odd.
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  3. #3
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    Re: infinite series

    Quote Originally Posted by HallsofIvy View Post
    The "general term" is a_n= \frac{1}{n/2}= \frac{2}{n} when when n is even, and (-1)\frac{2}{n-1} when n is odd.
    i still do not understand. how did you find a_n= \frac{1}{n/2}= \frac{2}{n} and (-1)\frac{2}{n-1} ?
    if that is the general term, then when n=1,then (-1)\frac{2}{n-1} = (-1)\frac{2}{0}? the first term when n=1 is should be  1

    maybe i should show how my book finds the sum the 2 series.

    for the 2., it's straightforward enough.
    S_n=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{n+1}
    \lim_ {n \to \infty} S_n=1
    therefore the series converges and  S=1

    for 1., it's different, and i wonder why they do not use the same method to get the sum and how exactly are the 2 series different.
    S_2_m_-_1=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+...+(-\frac{1}{m}+\frac{1}{m})=1
    S_2_m=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+...+(-\frac{1}{m}+\frac{1}{m})-\frac{1}{m+1}=1-\frac{1}{m+1}
    \lim_ {n \to \infty}S_2_m_-_1=\lim_ {n \to \infty}S_2_m=1
    therefore, the series converges and S=1
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  4. #4
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    Re: infinite series

    i think i get it now.


    for (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n+1})+...,
    the first term is (1-\frac{1}{2}), second term is (\frac{1}{2}-\frac{1}{3}), and so on
    sum would be 1-\frac{1}{n+1}


    for 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-...,
    the first term is 1, second term is -\frac{1}{2}, third term is \frac{1}{2} and so on
    as it approaches infinity, there could be and odd number of terms or an even number of terms.
    it isnt like the other series where every term come with pair of numbers and the numbers in the series surely consistently cancel each other out.
    if there's an even number of terms, it would be like the other series, but if there's an odd number of terms, there will be less or more 1 term in comparison with the other series.
    S_2_m_-_1 shows how the series would look if there's an odd number of terms and  S_2_m shows how it would be if there's an even number of terms

    i could be wrong or inaccurate, this is just how i understand it
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