$\sum_{n>k}\frac{j}{j+1}\cdot (j+1)^{-n}=\frac{j}{j+1}\cdot (j+1)^{-(k+1)}\cdot \sum _{n=0}^{\infty}(j+1)^{-n}$

Question

Given $1\leq j\leq 6$ and $k\in \mathbb{N}$, why is this equality holds?

$\sum_{n>k}\frac{j}{j+1}\cdot (j+1)^{-n}=\frac{j}{j+1}\cdot (j+1)^{-(k+1)}\cdot \sum _{n=0}^{\infty}(j+1)^{-n}$

Answer 1

$\sum_{n>k}\frac{j}{j+1}\cdot (j+1)^{-n}=\frac{j}{j+1}\sum_{n>k}(j+1)^{-n} =$

$=\frac{j}{j+1}\cdot \Big( (j+1)^{-k-1}+(j+1)^{-k-2}+(j+1)^{-k-3}+... \Big) =$

$=\frac{j}{j+1} \cdot (j+1)^{-k-1} \Big(1+(j+1)^{-1}+(j+1)^{-2}+... \Big)=$

$=\frac{j}{j+1} \cdot (j+1)^{-k-1} \sum _{n=0}^{\infty}(j+1)^{-n}$