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.
And then I'm stuck... I'm not sure how to go from the absolute value diverging to the series diverging. It clearly does not converge absolutely but what about conditional convergence? Please help!