A hint or two may suffice with this one.
For any arithmetic progression, which always approaches infinity as .
Therefore any AP is divergent.
Try to show that P(k+1) will definately be true "if" P(k) is true.
Therefore, write P(k+1) in terms of P(k) in order to draw a comparison.
"If" P(k) is true, then
Hence, if P(k) is true, then P(k+1) is true, since
Now all you need do is prove P(k) is true for the first relevant value of "n".