The geometric series:

sum_n=1^oo[r^n-1]

the general partial sum of the series

s_n=1+r+r²...r^n-1

rs_n=r+r²+r³...rⁿ

subtracting gives

s_n-rs_n=1-rⁿ

rearanging gives

s_n=(1-rⁿ)/(1-r)=1/(1-r)-rⁿ/(1-r)

so the sum of the series for |r|<1

^lim_n~oo[s_n]=1/(1-r)-^lim_n~oo[rⁿ/(1-r)]=1/(1-r)

sum_n=1^oo[r^n-1]=1/(1-r)