# Is it true that $\prod_{i=1}^{\infty} a_{i,t} \rightarrow \prod_{i=1}^{\infty}L_i$ when $a_{i,t} \rightarrow L_i$ for every $i$, and $L_i\to 1$?

Let $A_1,A_2,A_3,\dots$ be a sequence of sequences where each $$A_i = a_{i,1},a_{i,2},a_{i,3},\dots$$

Each sequence $A_i$ converges and in particular as $t \rightarrow \infty$, $a_{i,t} \rightarrow L_i$ for every $i$. We also have that in the limit as $i \rightarrow \infty$ the limits of these sequences converge to 1. I.e. as $i \rightarrow \infty$, $L_i \rightarrow 1$.

QUESTION: When does the following occur?

In the limit as $t \rightarrow \infty$ $$\prod_{i=1}^{\infty} a_{i,t} \rightarrow \prod_{i=1}^{\infty}L_i$$

I know how to prove that the limit of the product of two (convergent) sequences is the product of the limit of those sequences. I cannot see whether this argument extends to infinite products. Also I am not sure how to use the fact that the $L_i$ converge to 1.

• Titles should be more descriptive than "does this converge"? There are many thousands of questions about convergence here; titles should help them be recognized.
– user147263
Commented Aug 4, 2014 at 15:34
• If the $L_i$ don't converge to $1$ then the sequence of partial products will diverge or go to zero. Anyway, the best answer to this question is to consider the sequence of logarithms of the partial products and then apply a theorem about interchange of summation signs.
– Ian
Commented Aug 4, 2014 at 16:24

Taking the theorem about limits of finite products for granted, you are asking the following: given $a_{i,j} \to L_i$ as $j \to \infty$ and $L_i \to 1$ as $i \to \infty$, when do we have

$$\lim_{j \to \infty} \lim_{k \to \infty} \prod_{i=1}^k a_{i,j} = \lim_{k \to \infty} \lim_{j \to \infty} \prod_{i=1}^k a_{i,j}?$$

This is a question about interchange of limits. The generic answer is that you cannot do it: some additional hypotheses are required. The usual approach for infinite products is to convert to a problem about infinite sums by taking logarithms:

$$\ln \left ( \lim_{j \to \infty} \lim_{k \to \infty} \prod_{i=1}^k a_{i,j} \right ) = \lim_{j \to \infty} \lim_{k \to \infty} \sum_{i=1}^k \ln \left ( a_{i,j} \right )$$

using continuity of $\ln$ and the usual logarithm properties. Now if we can interchange the limits on the right side, then we can "undo" the logarithm procedure (that is, pass the $\ln$ back through the limits and apply $\exp$) to get the result.

The "advanced calculus criterion" for being able to interchange these limits is that the sum converges uniformly in $j$, that is, for any $\varepsilon > 0$ there exists $N > 0$ so that for all $n \geq N$ and all $j$

$$\left | \sum_{i=n}^\infty \ln \left ( a_{i,j} \right ) \right | < \varepsilon$$

One can prove that this condition is necessary. The cleanest proof I can think of rephrases summation in terms of integration with respect to the counting measure and then applies the Vitali convergence theorem. Then uniform integrability is trivial for summation, and tightness is equivalent to uniform convergence of the sums.

The condition that $L_i \to 1$ is required for us to use this procedure, because an infinite product can "diverge to zero" if all but finitely many terms are bounded a fixed distance below $1$. We say that it diverges to zero because in this case the infinite sum of the logarithms diverges to $-\infty$. In this case the sequence of partial products converges but the sequence of partial sums of logarithms does not. By assuming $L_i \to 1$ we remove this issue.

Usually this is not true. You must take care when consider the limit of a limit thing.

Here is a counterexample.

We define $a_{i,t}$ inductively. Let $a_{1,1}=1$ and $a_{1,t}=2$ if $t>1$. Assume that we defined $a_{j,t}$ for all $j<i$. Then we define $$a_{i,t}= \begin{cases} 1 &t<i\\ \prod_{j<i}a_{j,i}&t=i\\ 1+\frac{1}{i}&t>i \end{cases}$$

Then $$L_i=1+\frac{1}{i},$$ and $$\prod_{i=1}^\infty a_{i,t}=1.$$