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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.

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Hint: differentiate $\sum x^{n+1}$ twice, and then multiply by $x$. – David Mitra Feb 3 '12 at 19:15
What have you tried? Do you know the ratio test? @David: A bit more than Guy was asking for. – anon Feb 3 '12 at 19:18
He actually said he wanted to compute the value of the series. – azarel Feb 3 '12 at 19:22
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 that – Guy Feb 3 '12 at 19:23
Just follow David hint. – azarel Feb 3 '12 at 19:24
up vote 9 down vote accepted

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 $|x|<1$.

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Thanks a lot, I appreciate it – Guy Feb 3 '12 at 19:31

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