[Dan.]

In particular, the 2^Kth partial sum of the harmonic series can be expressed as S = 1 + K/2. So in this case, we allow P = 2^K. Following, K = ln (P - 2).

Therefore, the partial sum can be expressed as S = 1 + [ln(P - 2)]/2. Then the entire expression can be rewritten as S - ln P < 1.

This leads to:
1 + [ln(P - 2)]/2 -ln P < 1.
[ln(P - 2)]/2 - ln P < 0
[ln(P - 2)]/2 < ln P
ln(P - 2) < 2*ln P
ln(P - 2) < ln (P^2)
(P - 2) < P^2
P < P^2 + 2.

Clearly, for P > 1, P^2 > P. So the statement 1 + 1/2 + 1/3 +...+ 1/P - ln P < 1 for all P > 1 holds. QED