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How can I find to what this sum converges to?

$$\sum _{n=1}^{\infty} n(n+1)x^n$$

I proved that it converges when

$$|x| < 1$$

but no idea how to find what it sums to.

  • 7
    Hint: differentiate $\sum x^{n+1}$ twice, and then multiply by $x$.2012-02-03
  • 0
    What have you tried? Do you know the ratio test? @David: A bit more than Guy was asking for.2012-02-03
  • 0
    He actually said he wanted to compute the value of the series.2012-02-03
  • 0
    I used different tests to prove it converges. I need to find the sum it converges to. I'm not sure how to go about doing that2012-02-03
  • 0
    Just follow David hint.2012-02-03

1 Answers 1

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Let $$f(x)=x^1+x^2+x^3+\cdots$$ Then, for $|x|<1$ $$ f'(x)=1+2x+3x^2+4x^3+\cdots $$ $$\eqalign{ f''(x)&=2\cdot1\cdot x^0+3\cdot 2\cdot x+4\cdot3\cdot x^2+\cdots\cr &=\sum_{n=1}^\infty (n+1)n x^{n-1}. } $$ So $$xf''(x)=\sum_{n=1}^\infty (n+1)n x^n.$$

But, for $|x|<1$, $$ f(x)={x\over 1-x} $$ $$f'(x)={d\over dx}{x\over 1-x}={1\over (1-x)^2}$$ $$f''(x)={2\over( 1-x)^3}$$

So $$ {2x\over( 1-x)^3} = \sum_{n=1}^\infty (n+1)n x^n,$$ for $|x|<1$.

  • 0
    Thanks a lot, I appreciate it2012-02-03