Why does $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots$?

Question

I apologize if this is a duplicate.

I was taught how to prove that $\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\dots$, and one of the steps was to write the equality: $$\int \dfrac{1}{1+x^2} \ dx = \int \sum\limits_{n=0}^\infty x^{2n}\cdot(-1)^n \ dx = \int 1-x^2+x^4-x^6+x^8\dots \ dx$$ Why does $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots$? I have no idea on how to proceed with this. Could someone please point me in the right direction? Thanks in advance.

Answer 1

First, use this basic fact from geometric series: $$ \frac1{1-x}=\sum_{n=0}^\infty x^n. $$ Make the substitution $-x^2$ for $x$ to obtain $$ \frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}. $$

EDIT: to elucidate the first bit, suppose have the infinite series $$ a+r\cdot a+r^2\cdot a+\ldots $$ where $|r|<1$. Let $L$ be this sum, supposing it exists; $|r|<1$ is actually a necessary and sufficient condition. Then $L-a=r\cdot L$ by construction, so we have $$ a=(1-r)L\implies L=\frac a{1-r}. $$ In our case, we have $a=1$, $r=x$.

Answer 2

$$1=1(1-x)+x(1-x)+x^2(1-x)+...x^n(1-x).$$

Just substitute $-x^2$ in for $x$.

Alternatively, simply multiply both sides of your equation by $1+x^2$ to yield $1=1$, proving equality.

Answer 3

A geometric series is of a series for $\dfrac{1}{1-r}$. The article in wikipedia (http://en.wikipedia.org/wiki/Geometric_series) shows the series for this: $$\dfrac{1}{1-r}\mbox{ at $r=-x^2$ is }\sum^\infty_{k=0}(-x^2)^k=\sum^\infty_{k=0}(-1)^kx^{2k}$$ This gives you your required series, which converges iff $|x^2|<1$. Alternatively, you could Taylor expand about some point $a$ to get the series for $\dfrac{1}{1+x^2}$ about $a$.