or more generally,
a1 + a2 + a3 + ... + an-1 + an + n+1 + ...
Since its in geometric progression, an/an-1 = an+1/an
Taking log of both sides:
ln(an/an-1) = ln(an+1/an)
ln(an) - ln(an-1) = ln(an+1) - ln(an)
which proves arithmetic progression of the logs of terms in a geometric...