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Finding infinite series formulae

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    This isn't my homework problem, but I'd like to know how to prove [tex]\sum _{ n=0 }^{ \infty }{ \frac { n }{ { \alpha }^{ n } } } =\quad \frac { \alpha }{ { (\alpha -1) }^{ 2 } }[/tex]

    I only know basic convergence tests (including the integral test), and that
    [tex]\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \alpha }^{ n } } } =\frac { 1 }{ 1-\alpha } [/tex]

    Also, can anyone list some tools to tackle these problems with?

    2. Relevant equations
    [tex]\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \alpha }^{ n } } } =\frac { 1 }{ 1-\alpha } [/tex]

    Edit: correction to geometric series equation: [tex]\forall |\alpha |<1\quad \sum _{ n=0 }^{ \infty }{ { \alpha }^{ n } } =\frac { 1 }{ 1-\alpha }[/tex]
     
    Last edited: Jun 18, 2012
  2. jcsd
  3. Jun 18, 2012 #2

    LCKurtz

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    Certainly, that equation isn't correct; try ##\alpha = \frac 1 2##. But to give a general answer to your question, when you have a sum similar to $$
    \sum_{n=0}^\infty nx^n$$which would be a geometric series if that ##n## weren't out in front, you want to think about derivatives. Start with$$
    f(x) = \sum_{n=0}^\infty x^{n} = \frac 1 {1-x}\hbox{ if }|x|<1$$Differentiate both sides of that:$$
    \frac 1 {(1-x)^2}= \sum_{n=0}^\infty nx^{n-1} = \frac 1 x\sum_{n=0}^\infty nx^n$$
    Multiply both sides by ##x## and you have$$
    \sum_{n=0}^\infty nx^n = \frac x {(1-x)^2}$$Then put in whatever ##x## you wish, as long as ##|x|<1##.
     
    Last edited: Jun 18, 2012
  4. Jun 18, 2012 #3
    For starters, your geometric series is not right... It should be [itex] \alpha^n [/itex], not [itex] 1/\alpha^n [/itex].

    This looks a lot like you might want to differentiate the geometric series with respect to alpha. It should be easy to convince yourself that this is legal.
     
  5. Jun 18, 2012 #4
    Thank you very much
     
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