What is the function associated with the following power series: $\sum_{n=2}^{\infty}x^n\;\;?$ What is the function associated with the following power series: 
$$\sum_{n=2}^{\infty}x^n$$
If the sum started at 0 (if $n=0$) the function would be $\frac{1}{1-x}$, but with $n=2$, I am a bit confused.
Please pardon my poor english.
 A: Note that 
$$
\sum_{n = 2}^\infty x^n =  \sum_{n = 0}^\infty x^n - 1 - x,
$$
and then apply the  formula for $\sum_{n = 0}^\infty x^n$.
A: We assume that $|x|<1$. By letting $n=m+2$, we have that
$$\sum_{n=2}^{\infty}x^n=\sum_{m=0}^{\infty}x^{m+2}=x^2\cdot\sum_{m=0}^{\infty}x^{m}=\frac{x^2}{1-x}$$
More generally, in the same way, we show the closed formula
$$\sum_{n=n_0}^{\infty}x^n=\frac{x^{n_0}}{1-x}.$$
A: $$ \sum_{n=2}^{\infty}x^n = x^2+x^3+x^4+...= x^2( 1+x+x^2+...)= \frac {x^2}{1-x}$$
Which converges for $|x|<1.$
A: Note that
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
is true for $|x|<1$, then note that
$$\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{k-1}x^n+\sum_{n=k}^{\infty}x^n =\frac{1}{1-x}\implies\sum_{n=k}^{\infty}x^n =\sum_{n=0}^{\infty}x^n-\sum_{n=0}^{k-1}x^n=\frac{1}{1-x}-\sum_{n=0}^{k-1}x^n$$
A: Assume that
$\vert x \vert < 1; \tag 1$
then
$\displaystyle \sum_2^\infty x^n = \sum_0^\infty x^n - (1 + x)$
$= \dfrac{1}{1 - x} - (1 + x) = \dfrac{1}{1 - x} - \dfrac{1 - x^2}{1 - x} = \dfrac{1 - (1 - x^2)}{1 - x} = \dfrac{x^2}{1 - x}. \tag 2$
