Dominated convergence for infinite product proof Dominated convergence theorem for an infinite product states that:  
$$\lim_{n \to ∞} \prod_{k=1}^{∞}(a_{kn}+1)=\prod_{k=1}^{∞}\lim_{n \to ∞} (a_{kn}+1)$$ 
If :
There exists a convergent sum $$\sum_{k=1}^{\infty}b_{k}$$ such that (for all k)$$b_{k}{\ge}|a_{nk}|$$ 
My question is : Why is the above theorem true ? (I think it has something to do with using the dominated convergence theorem  for infinite 
sums but I don’t know how.)
 A: Hint: first recall that the logarithm is a continuos function in its natural domain of definition. Thus, the logarithm of a limit is the limit of the logarithm. Now use this fact together with the well known property:
$$\ln(ab)=\ln(a) + \ln(b)$$
($a,b>0$), which can be easily extended to infinite products.
EDIT: suppose $a_{nk} >0$ (otherwise, you will just need to change signs). Then, the logarithm is positive and you get:
$$e^{b_k} -1 \geq a_{nk}$$
Thus, you have found
$$c_k := e^{b_k} - 1$$
which is you what you were looking for.
You can proceed similarly for the other cases.
EDIT: You could proceed as in Under what conditions can we swap a limit with an infinite product?. If you assume that the absolute value of the logarithm is bounded, you have concluded.
EDIT: I will rephrase the proof of the link above, applying it to your case. Suppose that there exist $c_k$ such that
$$ |\ln(a_{nk} + 1)| \leq c_k $$
with $\sum_{k} c_k$ convergent. Then, by the Dominated Convergence Theorem for series we have:
$$ \lim_{n \rightarrow \infty} \sum_{k} \ln(a_{nk} +1) = \sum_{k} \lim_{n \rightarrow \infty} \ln(a_{ nk} +1) $$
Notice that the inequality for the logarithm implies the following (let $d_{nk}:= a_{nk} + 1 > 0$):
$$ |d_{ nk}| \leq e^{c_k} $$
But then:
$$ \prod_{k} e^{c_k} = e^{\sum_{k} c_k} $$
is convergent. Moreover, you can easily see that you can interchange limit and infinite product by the equation above involving the sums of logarithms. Thus, when there is a bound $c_k$ as above for the logarithms, the dominated convergence theorem also holds for infinite products.
