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Thread: another sum of the series

  1. #1
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    another sum of the series

    How do I find the sum of this series?

    summation of (n+1)/3^n

    when I used the first few n's I didn't see a pattern.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    How do I find the sum of this series?

    summation of (n+1)/3^n

    when I used the first few n's I didn't see a pattern.
    It is the combination of arithmetic and geometric series.
    Let $\displaystyle S=\sum \frac{n+1}{3^n}$
    Find 3S
    Find 3S-S

    KeepSmiling
    Malay
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  3. #3
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    Can you explain a little more? Still lost
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  4. #4
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    Hello, Nichelle14!

    How do I find the sum of this series? .$\displaystyle \sum^{\infty}_{n=1} \frac{n+1}{3^n}$

    We are given: . . $\displaystyle S \;= \;\frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \,\cdots$

    Multiply by $\displaystyle \frac{1}{3}:\;\;\frac{1}{3}S \;= \qquad\;\frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \,\cdots$

    Subtract: . . . . $\displaystyle \frac{2}{3}S \;= \;\frac{2}{3} + \underbrace{\frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} +\,\cdots}$

    After the first term, we have a geometric series with first term $\displaystyle a = \frac{1}{3^2}$
    . . common ratio $\displaystyle r = \frac{1}{3}$, and whose sum is: $\displaystyle \frac{\frac{1}{3^2}}{1 - \frac{1}{3}}\:=\:\frac{1}{6}$

    Hence, we have: .$\displaystyle \frac{2}{3}S\:=\:\frac{2}{3} + \frac{1}{6} \:=\:\frac{5}{6}$

    Therefore: .$\displaystyle \boxed{S\:=\:\frac{5}{4}}$

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