# \$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.

## Answer

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\$.

EDIT:

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}\$\$
Therefore:

\$\$s_n-xs_n=1-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}\$\$