Taylor series about (3+x)/(x+4)*exp(-x) expanded at x = - 4. How do i replicate what I see in wolfram? I'm puzzled by wolfram alpha's results.   If i ask 'series $\frac{3+x}{4+x} (\exp  (-x))$ expanded at -4' It will return a series.   What method does it use to do this?   I'm familiar with Taylor series but doing this will lead to having zero in the denominator when -4 is plugged in. Anyone familiar with series expansions able to help?
Here is wolfram alpha's results:
$\frac{1}{20} e^4 (x+4)^4-\frac{5}{24} e^4 (x+4)^3+\frac{2}{3} e^4 (x+4)^2-\frac{3}{2} e^4 (x+4)-\frac{e^4}{x+4}+2 e^4$  
(Where it is for some reason listed in reverse order when copied out of the program.)
Here is a link to the wolfram alpha output 
Even if i plug in values of n for the bottom series it doesnt seem to work or i just dont understand how to read this series.  Can someone please help walk me through this?
$\sum _{n\geq 0} \frac{(-1)^n e^4 (4+x)^n (n!+(1+n)!)}{n! (1+n)!}+\sum _{1+n=0} -e^4 (4+x)^n$
 A: Note that the function has a pole at (x+4). 
One way to get around this is to first multiply by $(x+4)$, take the taylor series as usual and then divide by $(x+4)$ term by term.
Another way is to isolate the root, by first multiplying by $(x+4)$ and setting $x=-4$. This gives $-e^4$
So subtract
$$
-\frac{e^4}{x+4}
$$
from the function. This will get rid of the pole. Proceed as usual and then add back
$$
-\frac{e^4}{x+4}
$$
By the way the $-e^4$ is called the residue of the pole.
I've done the work by hand and typed it up in LaTeX.  Here it is for anyone to see.
Ok so the first method takes me through the following steps:
$f^{0}(x) = (3+x)e^{-x},$  $f^{0}(-4) = -e^{4}$
$f^{1}(x) = -(2+x)e^{-x}+e^{-x},$  $f^{1}(-4) = 2e^{4}$
$f^{2}(x) = (1+x)e^{-x},$  $f^{1}(-4) = 3e^{4}$
so then to approximate with a quadratic $f(x) = -e^{4}+2e^{4}(x+4)-\frac{3e^{4}(x+4)^{2}}{2}$
now divide through with (x+4)
$f(x) = \frac{-e^{4}}{4+x}+\frac{2e^{4}(x+4)}{4+x}-\frac{3e^{4}(x+4)^{2}}{2(x+4)}$
you can see the simplification from here. Hope this helps.
