Question

how to prove
\[\frac1{1-n}=1+n+n^2+...+n^\infty=\sum_{i=0}^\infty n^i\]
for \(n<0\)

Answer 1

Sum in GP is given by -\[\frac{a(1-r^{n})}{1-r}\]
putting n = infinity
\[\frac{a(1-r^{\infty})}{1-r}\]

\[=\frac{a}{1-r}\]

Answer 2

you can't prove it for n<0

Answer 3

only holds for
|n| < 1

Answer 4

\[A=1+n+n^2+...+n^m\]\[nA=n+n^2+n^3+...+n^{m+1}\]\[nA-A=n^{m+1}-1\]\(n\neq1\)\[\Rightarrow A=\frac{1-n^{m+1}}{1-n}\]now if \(|n|<1\) as \(m \rightarrow \infty\)\[A=\frac{1}{1-n}\]