Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges
The Attempt at a Solution
So far I have this:
Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then [tex]\exists[/tex] N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)
the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).
Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.
Then I'm stuck. I don't know how to get from the absolute value diverging to the series diverging. It obvioulsy does not converge absolutely, but what about conditional convergence. Please help!