You need to show . Due to your remark, this can be rewritten as
where if and if .
You need a theorem to justify that. A good one would be the "dominated convergence theorem" (you may know another name):Dominated convergence theorem (for series):I wrote the theorem for series, so you must let in order to have and apply the theorem.
If is such that
- for every , ;
- there is such that for every , and ,
Then, for every , the series and converge, and we have .
Thus you must find such that converges and, for every ,
.(And the inequality is obvious if since we would have )
To this aim, you can use the following bounds: if , then . I let you carry on from there.
Note: I assumed so that the factors of the product are positive and the logarithm makes sense.