Question

Prove that for any constant a, |r|<1, then sum ar^k=a/(1-r) for k from 0 to infinity

Answer 1

the sum of a geometric progression is a(1-r^n)/(1-r)

Answer 2

Typing these proofs out in latex takes so long... X_X

First we'll start with:
\[a+ar+ar^2+ar^3+...\]
\[= \lim_{n \rightarrow \infty}(a+ar+ar^2+ar^3+ ... +ar^n)\]
Now we can use the formula for geometric progression:
\[a+ar+ar^2+ar^3+...+ar^n = a(1 - r^{n+1})/(1 - r)\]
and then we get:
\[= \lim_{n \rightarrow \infty}(a(1 - r^{n+1})/(1 - r))\]
\[r^{n+1} \rightarrow 0\space for\space|r| < 1\]
\[\therefore\space=a/(1-r)\]

Answer 3

as r < 1 r^inf -> 0 therefore you get a*1/(1-r)