Solving a sum using Cauchy's sequence

Question

could I get some help proving that the sum of this expression is equal to $m$ using Cauchy's criterion?

$ 1+\frac{m-1}{m}+(\frac{m-1}{m})^{2}+(\frac{m-1}{m})^{3}+(\frac{m-1}{m})^{4}+....+(\frac{m-1}{m})^{n} $

Answer 1

With geometric series:

$1+\frac{m-1}{m}+(\frac{m-1}{m})^{2}+(\frac{m-1}{m})^{3}+(\frac{m-1}{m})^{4}+....+(\frac{m-1}{m})^{n} \to \frac{1}{1-\frac{m-1}{m}}=m$ as $ n \to \infty.$