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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.

  • 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