the non oscilating geometric series described in this problem may be considered the values of the function:
f(x) = ar^(x-n)

for discrete values of x
Where of course the first value of x would be equal to n

f(x) = a(r^-n)(r^x)
f'(x)= a(r^-n)ln(r)r^x
f"(x)=a(r^-n)[ln(r) * ln(r)] r^x
as f(x) is non oscillating, r is positive, and r^(any power) must necessarily be positive
as f(x) by our difinition is greater than zero, a will be greater than zero ie positive

ln(r) will be negative for r
hence f"(x) is always positive (ie function is concave up)
meaning between any two points f(a) and f(b), the value of the function f(x) will lie below the straight line joining f(a) and f(b)

The arithmatic sum IS the value of g(x) where g(x) is the equation of the straight line joining f(a) and f(b)


each term of arithmatic sequence is greater than or equal to its corresponding term in the geometric sequence.
hence the arithmatic sum will always be greater than or equal to the geometric sum (equal to when a +1 = b)

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