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Math Help - Sum of Series

  1. #1
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    Sum of Series

    Find the sum of the series \sum_{n=0}^{\infty} \frac{3n+5}{4n}

    I know that the limit of the function is also the sum of the series but I got 0 for the limit which I'm not so sure if it is right
    can anyone help me and see if thats correct?
    Last edited by JC05; April 28th 2012 at 11:15 PM.
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    Find the sum of the series \sum_{n=0}^{\infty} \frac{3n+5}{4n}

    I know that the limit of the function is also the sum of the series but I got 0 for the limit which I'm not so sure if it is right
    can anyone help me and see if thats correct?
    No, the limit of the function is NOT the value of the sum. The reason you need to work out this limit is because a series will never be convergent if the limit of the function is not 0.

    It's pretty easy to see that

    \displaystyle \begin{align*} \lim_{n \to \infty}\frac{3n + 5}{4n} &= \lim_{n \to \infty}\left(\frac{3}{4} + \frac{5}{4n}\right) \\ &= \frac{3}{4} \\ &\neq 0 \end{align*}

    The series is divergent.
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  3. #3
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    Re: Sum of Series

    So is the sum of the series 3/4
    I kinda feel so stupid now ... I should've seen that
    Thanks though!!!
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    So is the sum of the series 3/4
    I kinda feel so stupid now ... I should've seen that
    Thanks though!!!
    No, the sum of the series is NOT 3/4. Since this limit is not 0, the series is DIVERGENT! It shoots off to infinity.
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    Re: Sum of Series

    inf-sigma-0 3^n/4^n + inf-sigma-0 5/4^n

    work from there.
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    Re: Sum of Series

    Quote Originally Posted by manhole View Post
    inf-sigma-0 3^n/4^n + inf-sigma-0 5/4^n

    work from there.
    Don't confuse the OP. He did not write \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}, he wrote \displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}, so your simplification is not valid.
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    Re: Sum of Series

    Quote Originally Posted by Prove It View Post
    Don't confuse the OP. He did not write \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}, he wrote \displaystyle \begin{align*} \frac{3n + 5}{4n} \end{align*}, so your simplification is not valid.
    MY BAD!!! it is \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}

    No wonder I was getting super confused ...
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    MY BAD!!! it is \displaystyle \begin{align*} \frac{3^n + 5}{4^n} \end{align*}

    No wonder I was getting super confused ...
    Well then do as you were instructed by by Manhole, his psychic powers must be a little more powerful and finely tuned than mine...

    \displaystyle \begin{align*} \sum_{n = 0}^{\infty}\frac{3^n + 5}{4^n} &= \sum_{n = 0}^{\infty}\left(\frac{3^n}{4^n} + \frac{5}{4^n}\right) \\ &= \sum_{n = 0}^{\infty}\left(\frac{3}{4}\right)^n + 5\sum_{n = 0}^{\infty}\left(\frac{1}{4}\right)^n \end{align*}

    The first is an infinite geometric series with \displaystyle \begin{align*} a = 1, r = \frac{3}{4} \end{align*} and the second is an infinite geometric series with \displaystyle \begin{align*} a = 1, r = \frac{1}{4} \end{align*}.
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    Re: Sum of Series

    Quote Originally Posted by Prove It View Post
    Well then do as you were instructed by by Manhole, his psychic powers must be a little more powerful and finely tuned than mine...

    \displaystyle \begin{align*} \sum_{n = 0}^{\infty}\frac{3^n + 5}{4^n} &= \sum_{n = 0}^{\infty}\left(\frac{3^n}{4^n} + \frac{5}{4^n}\right) \\ &= \sum_{n = 0}^{\infty}\left(\frac{3}{4}\right)^n + 5\sum_{n = 0}^{\infty}\left(\frac{1}{4}\right)^n \end{align*}

    The first is an infinite geometric series with \displaystyle \begin{align*} a = 1, r = \frac{3}{4} \end{align*} and the second is an infinite geometric series with \displaystyle \begin{align*} a = 1, r = \frac{1}{4} \end{align*}.
    Its ok.. apparently manhole is in my class
    So i think i solved the equation which came out to be 10.33
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    Its ok.. apparently manhole is in my class
    So i think i solved the equation which came out to be 10.33
    I disagree with your answer.
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    Re: Sum of Series

    Quote Originally Posted by Prove It View Post
    I disagree with your answer.
    Why...?
    Ok so the first thing i did was calculate \frac{1}{1-\frac{3}{4}} which equal 4

    then i did the second one \frac{1}{1-\frac{1}{4}} which equal 1.33 and then i added 5 to it which makes 6.33

    Then i combined both equation to get 10.33
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  12. #12
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    Why...?
    Ok so the first thing i did was calculate \frac{1}{1-\frac{3}{4}} which equal 4

    then i did the second one \frac{1}{1-\frac{1}{4}} which equal 1.33 and then i added 5 to it which makes 6.33

    Then i combined both equation to get 10.33
    You're not supposed to add 5 to the second one, it's supposed to be MULTIPLIED by 5.
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    Re: Sum of Series

    Quote Originally Posted by Prove It View Post
    You're not supposed to add 5 to the second one, it's supposed to be MULTIPLIED by 5.
    Sorry....
    10.67
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    Re: Sum of Series

    Quote Originally Posted by JC05 View Post
    Sorry....
    10.67
    That's better.

    Of course, we should always stay exact where possible, so it would be better if you keep the answer as \displaystyle \begin{align*} \frac{32}{3} \end{align*}.
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