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Math Help - What is the sum as n approaches infinity?

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    What is the sum as n approaches infinity?

    What is the sum as n approaches infinity of 5(\frac{1}{3})^{r - 1}? Sorry I don't know how to write the Summation sign with the infinity symbol at the top and the r = 1 at the bottom of the summation sign in LaTeX.
    Last edited by mark1950; September 7th 2009 at 05:23 AM.
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    Hello mark1950
    Quote Originally Posted by mark1950 View Post
    What is the sum as n approaches infinity of 5(\frac{1}{3})^{r - 1}? Sorry I don't know how to write the Summation sign with the infinity symbol at the top and the r = 1 at the bottom of the summation sign in LaTeX.
    You want to know the answer to the summation S=\sum_{r=1}^{\infty}5\times\Big(\frac{1}{3}\Big)^  {r-1}.

    Let's put the first few values of r in:

    S = 5\times\Big(\frac{1}{3}\Big)^{0}+5\times\Big(\frac  {1}{3}\Big)^{1}+5\times\Big(\frac{1}{3}\Big)^{2}+.  ..

    = 5 + 5\times\Big(\frac{1}{3}\Big)^{1}+5\times\Big(\frac  {1}{3}\Big)^{2}+...

    which you may now recognise as a Geometric Progression (or Series) whose first term, a= 5, and common ratio r=\frac{1}{3}. The sum to infinity of such a series is \frac{a}{1-r}. So in this case:

    S = \frac{5}{1-\frac13}=\frac{5}{\tfrac23}=\frac{15}{2}=7.5

    Grandad
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    How about if the progressions aren't geometric? Like,

    \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}

    and

    \sum_{r=1}^{\infty}(r + 1)(r + 2^r) from the progressions, 2.3 + 3.6 + 4.11 + 5.64 + ...
    Last edited by mark1950; September 7th 2009 at 06:11 AM.
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  4. #4
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    Hello mark1950
    Quote Originally Posted by mark1950 View Post
    How about if the progressions aren't geometric? Like,

    \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}

    and you have to find the sum to infinity?
    There are no hard and fast rules, because there are so many variations. But in this case, you use Partial Fractions, and say that

    \frac{1}{(r+1)(r+2)}= \frac{1}{r+1}-\frac{1}{r+2}

    so that

    \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\sum_{r=1}^{\infty}\Big(\frac{1}{r + 1}-\frac{1}{r + 2}\Big)

    =\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=1}^{\infty}\frac{1}{r + 2}

    =\sum_{r=1}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}

    =\sum_{r=1}^{1}\frac{1}{r + 1}+\sum_{r=2}^{\infty}\frac{1}{r + 1}-\sum_{r=2}^{\infty}\frac{1}{r + 1}

    =\frac12

    Grandad
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    Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.
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    Quote Originally Posted by mark1950 View Post
    Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.
    Hi

    I think you can do it using the method of differences but it is longer than Grandad's solution .

    Same thing , do the partial fraction first .

    =\sum^n_{r=1}[\frac{1}{r+1}-\frac{1}{r+2}]

    =\sum^n_{r=1}[f(r)-f(r+1)]

    =f(1)-f(n+1)

    =\frac{1}{2}-\frac{1}{n+2}

    =\frac{n+1}{2n+4}

    Then \sum_{r=1}^{\infty}\frac{1}{(r + 1)(r + 2)}=\lim_{n\rightarrow\infty}\frac{n+1}{2n+4}

    =\frac{\frac{n}{n}+\frac{1}{n}}{\frac{2n}{n}+\frac  {4}{n}}

    =1/2
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  7. #7
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    Hello mark1950
    Quote Originally Posted by mark1950 View Post
    Is it possible if I use the method of difference? If I used it, how would it looked like? Sorry for my inquiry.
    That is essentially the same as the method I've used here. It's just that instead of writing individual terms like:

    \frac12-\frac13+\frac13-\frac14+\frac14-\frac15+...

    I've left the series as two summations using the
    \Sigma sign, and fiddled the limits around. It's just a bit neater that way.

    Grandad
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