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.


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?


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 $0$.

  • $\begingroup$ 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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