- If S[n]= 1 + r + r^2 + ... +r^n,

then rS[n] = r + r^2 + ... +r^n + r^(n+1) and if we subtract, then most of the terms cancel,

so S[n]-rS[n] = 1 - r^(n+1)

and so S[n] = (1- r^(n+1))/(1-r), or if you prefer (r^(n+1)-1)/(r-1).

So S[n] = (1)/(1-r) - (r^(n+1))/(1-r)

If |r| < 1 then large powers of r will get very small and so the second term will go to zero as n goes to infinity. So in that case the infinite sum is just 1/(1-r).

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