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I have to find a power series representation and interval of convergence for $$f(x) = \frac{x-x^2}{(1+2x)^3}$$

Noting that $\frac{1}{1+2x}=\frac{1}{1-(-2x)}=\sum_{n=0}^\infty(-2x)^n$, I start taking derivatives until I arrive at $$\frac{1}{(1+2x)^3} = \frac{1}{8}\sum_{n=0}^\infty (-1)^n(2)^{n+2}(n+2)(n+1)x^n = \sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n$$

But I'm looking for $$\frac{x-x^2}{(1+2x)^3} = (x-x^2)\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n$$

This is where I'm stuck. In general, I just don't know how to deal with this sort of situation where I have to multiply a power series by multiple separate values in the function's numerator. The manipulations required on the separate sums trip me up. For example, in this case I tried

$$\left(x\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n\right) - \left(x^2\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n\right)$$

Which eventually gives me

$$\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)\left(x^{n+1} - x^{n+2}\right) = \sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)\left(x^n(x - x^2)\right) $$

Is this right? It really seems too complicated to me, I can't see any simple way to apply the ratio test to this so as to obtain its interval of convergence.

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Hint. Once you arrive at $$ x\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n=\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^{n+1} $$ you can make a change of index, setting $m=n+1$ thus $n=m-1$, giving $$ \sum_{n=0}^\infty (-1)^n2^{n-1}(n+2)(n+1)x^{n+1}=\sum_{m=1}^\infty (-1)^{m-1}2^{m-2}(m+1)mx^m $$ then collecting the general terms in the two series.

Can you take it from here?

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Oh, you're almost there...

What you seem to miss here is reindexing.

Group the (infinite) sums you got according to $x^n$'s:

$$\sum_{n\ge0} a_nx^{n+1} + \sum_{n\ge0} b_nx^{n+2} = \sum_{k\ge 1}(a_{k-1}+b_{k-2})x^k $$ where $k=n+1$ is the new index, and $b_{-1}$ is taken as $0$.

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  • $\begingroup$ I studied up a bit on reindexing and tried this. I ultimately arrived at $$x+\sum_{k=2}^{\infty}[(-1)^{k-1}(2)^{k-2}(k+1)(k)-(-1)^{k-2}(2)^{k-3}(k)(k-1)]x^k$$, which looks similar to your final sum $$\sum_{k\geq 1}(a_{k-1} + b_{k-2})x^k$$ Is this correct? If so, would I just use the ratio test as normal to find the interval of convergence for it? $\endgroup$ – enharmonics Apr 25 '16 at 21:06
  • $\begingroup$ Well, yes it should be something like that. $\endgroup$ – Berci Apr 26 '16 at 19:32

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