$f(x)={1over (x-1)(x+2)}$ Taylor series.

Find Taylor series around $x_0=0$ for: $$f(x)={1over (x-1)(x+2)}=(text{By a hintby simple algebra}){1over 3}left[{1over x-1}-{1over x+2}right]$$. Check where the series converges to the function $f$.
I am really weak in this kind of questions. Thorough explanation are more then welcome. Short explanation and hints are welcome as well.


Since $$sum_{n=0}^infty x^n ={1 over 1-x} , |x|<1 $$

We have: $$-sum_{n=0}^infty x^n ={1 over x-1} , |x|<1$$

$${1 over x+2}={1 over 2}{1 over 1-(-x/2)}={1 over 2}sum_{n=0}^infty left (-frac{x}{2} right )^nquad , |x|<2$$

Since the series of the sum is the sum of the series we have:

$$f(x)={1 over 3} left [ {1 over x-1} – {1 over x+2}right ]= {1 over 3} left [ -sum_{n=0}^infty x^n – {1 over 2} sum_{n=0}^infty left (-frac{x}{2} right )^nright ]=sum_{n=0}^infty left [-{1 over 3} left ( 1 +{1 over 2}left (-{1 over 2}right )^nright ) right ] x^n$$

for $|x|<1$.


The first formula is the geometric series.

Let $x in Bbb{R}$ with $|x|<1$. Put $s_n=1+x+x^2+cdots +x^n$, note that:

$$x s_n=x+x^2+cdots +x^{n+1}$$


$$s_n={1-x^{n+1}over 1-x}$$

Since $|x|<1$, $x^n rightarrow 0$ as $n rightarrow +infty$, which gives:

$$lim_{nrightarrow infty}s_n=lim_{nrightarrow infty}{1-x^{n+1}over 1-x}={1 over 1-x}$$

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