Find power series using long division How do you find the power series of $\frac{1}{2+x}$ using long division?
When I do the long division, I get zero with a remainder of 1, giving me the original function back.
 A: It's not the usual kind of division where you eliminate high powers of $x$ and work your way down. Instead, you eliminate low powers of $x$ and work your way up (indefinitely). 
It's a bit hard to write a division out in Latex, but the idea is this.  You're dividing $1$ by $2 + x$. $2$ goes into $1$, $1/2$ times, so the first term in the series is $1/2$. We have a remainder of $1 - (1/2)(2 + x) = - (1/2)x$. 
Next, $2$ goes into $-(1/2)x$, $-(1/4)x$ times. So the second term in the series is $-(1/4)x$. The remainder now is $-(1/2)x + [(1/4)x](2 + x) = (1/4)x^2$.
$2$ goes into $(1/4)x^2$, $(1/8)x^2$ times. So the next term in the series is $(1/8)x^2$.
Carry on like this. More information on this method can be found here.
A: To understand this you have to understand how long division works. Long division (of $a/b$) is in fact a sequence of rewriting the division on the form:
$$a = qb + r$$
starting with $q=0$ and $r=a$ and successively rewrite that with different $q$s and $r$s to minimize $r$ (in some sense).
So now we would do that for $1/(2+x)$ so we start with
$$1 = 0(2+x) + 1$$
next we're depending on whether $x$ is large or small, if you want a normal power series you would require $x$ to be small (which makes the $2$ more important):
$$1 = {1\over 2}(2+x) + 1 -{1\over2}(2+x) = {1\over2}(2+x) - {1\over2}x$$
$$1 = \left({1\over 2} - {1\over4}x\right)(2+x) - {1\over 2}x + {1\over4}x(2+x) = \left({1\over2}-{1\over4}x\right)(2+x) + {1\over4}x^2$$
etc
further steps will be:
$$1 = \left(-\sum_0^{n-1} (-2)^{-k-1}x^k\right)(2+x) + (-2)^nx^n$$
If you on the other hand decide that $x$ is large you will end up with a Laurent series (a power series in $1/x$) since we will be striving to reduce the highest power of $x$ instead of the lowest:
$$1 = 0(2+x)+1$$
$$1 = x^{-1}(2+x) + 1 -x^{-1}(2+x) = x^{-1}(2+x) - 2x^{-1}$$
$$1 = (x^{-1} - 2x^{-2})(2+x) - 2x^{-1} + 2x^{-2}(2+x) = (x^{-1} - 2x^{-2})(2+x) + 4x^{-2}$$
etc
further steps will be
$$1 = \left(\sum_1^n (-2)^{k-1}x^{-k}\right)(2+x) + (-2)^{n}x^{-n}$$
The distinction between $x$ being large or small is (of course) whether $|x|>2$ or $|x<2|$. For $x=2$ you shouldn't expect any of the series to converge and for $x=-2$ you would have division by zero.
A: HINT: 
Rewrite $\frac{1}{2+x}$ as $(x+2)^{-1}$ and use Negative Binomial Theorem.
You will get $$(x+2)^{-1}=\sum_{k=0}^\infty (-1)^k \binom{k}{k}x^k\cdot 2^{-1-k} $$
A: $$\dfrac1{2+x}=\dfrac12\cdot\dfrac1{1+\dfrac x2}$$
Now use long division OR apply generalized Binomial Theorem  
$$\dfrac1{1+\dfrac x2}=\left(1+\dfrac x2\right)^{-1}$$
