How do you express $\dfrac 1{1+x}$ as an infinite polynomial? I have the function $\dfrac 1{1+x}$ which I want to express as an infinite polynomial. I believe the correct term is Taylor Series. How do I solve this problem?
 A: Use the geometric series
$$\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{i=0}^n (-x)^n\quad\text{for}\quad|x|<1$$
A: Brute force Taylor anyone?
$$f^{(n)}(x)=\frac{d^n}{dx^n}\frac1{1+x}=\frac{n!(-1)^n}{(1+x)^{n+1}}$$
If you want a proof, use induction.
At $x=0$, we get
$$=n!(-1)^n$$
Putting this into Taylor's theorem:
$$\frac1{1+x}=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{n!(-1)^n}{n!}x^n$$

$$\frac1{1+x}=\sum_{n=0}^\infty(-1)^nx^n$$

A: Because $1 - x + x^2 - x^3 + x^4 - \cdots = \dfrac {1}{1+x}$
It is sum of geometric series aka infinite polynomial.
A: Let's assume such a polynomial exist so that $\frac 1{1+x} = \sum a_i x^i$
Then $1 = (1+x)\frac1{1+x} = \sum a_i x^i + \sum a_i x^{i+1} = a_0 + \sum (a_{i+ 1} + a_i)x^i$.
So $a_0 = 1$ and $a_{i} + a_{i-1} = 0; a_{i} = -a_{i-1} \forall i > 0$.
Inductively that means $a_i = 1$ if $i$ is even and $a_i = -1$ if $i $ is odd.  Or $a_i = (-1)^i$.
So the polynomial is $\sum_{i=0}^{\infty}(-1)^ix^i$.  This is the geometric series one learns about in calculus.  It does require a caveat that unless $|x| < 1$ the infinite sum will not converge.
