What is the limit of this infinite sum of an infinite product? In a statistics problem (the Ross-Littlewood paradox) we encounter the following sum term
$$\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)$$
how do we evaluate such term?
Could we use the following with a limit on both sum and product together
$$\lim_{l \to \infty} \sum_{k=1}^l \prod_{n=k}^l \left( \frac{9n}{9n+1} \right) = \lim_{l \to \infty} \frac{9l}{10} = \infty$$
Or should we evaluate the terms independently
$$\lim_{l \to \infty} \sum_{k=1}^l \lim_{m \to \infty} \prod_{n=k}^m \left( \frac{9n}{9n+1} \right) = \lim_{l \to \infty} \sum_{k=1}^l 0 = 0$$

I have a visual interpretation of these integrals which is like a triangular form: 
$$\begin{array}\\
\sum_{k=1}^l \prod_{n=k}^m \left( \frac{9n}{9n+1} \right) =   & \frac{9 \cdot 1}{9 \cdot 1 +1} & \cdot & \frac{9 \cdot 2}{9 \cdot 2 +1} & \cdot & \frac{9 \cdot 3}{9 \cdot 3 +1} & \cdots & \cdots & \cdots & \cdots & \frac{9 \cdot m}{9 \cdot m +1} & + &  & \\ 
& &   & \frac{9 \cdot 2}{9 \cdot 2 +1}  & \cdot & \frac{9 \cdot 3}{9 \cdot 3 +1}  & \cdots & \cdots & \cdots& \cdots &  \frac{9 \cdot m}{9 \cdot m +1} & +\\ 
& &  &  &  & \frac{9 \cdot 3}{9 \cdot 3 +1}  & \cdots & \cdots & \cdots & \cdots  & \frac{9 \cdot m}{9 \cdot m +1} &+\\ 
& &  &  &  &  & \ddots & \cdots & \cdots & \cdots & \frac{9 \cdot m}{9 \cdot m +1} & +\\ 
& & &  &  &  &  & \ddots & \frac{9 \cdot l}{9 \cdot l +1} & \cdots & \frac{9 \cdot m}{9 \cdot m +1} \\ 
\end{array}  $$ 
and when ${(l,m) \to (\infty,\infty)}$ I suspect we end up with the same triangle independent from the path.

Of particular interest is the reasoning on the difference between the two. 


*

*Why it is a different result while the terms for both limits count up to the same collection? 

*Does the infinite term in the second case make sense. Would it be valid to state that $lim_{k \to \infty} \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right) = 0$ while this result has only been shown for finite $k$? 

*Is the product term zero or almost zero? 

*Is there literature on this topic of such nested expressions?

 A: If we put $$a_k=\prod_{n=k}^\infty \left(\frac{9n}{9n+1}\right)$$
then $a_k=0$ for every $k$, then
$$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty 0=0$$
Which agrees with your second reasoning, it seems more intuitive that the infinities aren't linked though I guess the notation might be a bit ambiguous.

Why it is a different result while the terms for both limits count up to the same collection?

Well the partial sum/product is clearly different with $\sum\limits_{k=1}^l\prod\limits_{n=k}^l$ then with $\sum\limits_{k=1}^l\prod\limits_{n=k}^m$, in the first case the value is completely determined with $l$ while in the second $l,m$ may vary.

Does the infinite term in the second case make sense. Would it be valid to state that $\lim_\limits{k \to \infty} \prod\limits_{n=k}^\infty \left( \frac{9n}{9n+1} \right) = 0$ while this result has only been shown for finite $k$?

Yes it's valid because you have that $\prod_\limits{n=k}^\infty\left(\frac{9n}{9n+1}\right)=0$ for every $k$ and $\lim_\limits{k\to\infty}0=0$

Is the product term zero or almost zero?

Product term is exactly zero, while for example $\frac1n$ when $n$ is large is almost zero we have that $\lim_\limits{n\to\infty}\frac1n=0$ which means exactly $0$

Is there literature on this topic of such nested expressions?

I couldn't quite find anything on this particular case but perhaps this might be of interest, it is about double sequences and double series. As a side note I think that the notation is ambiguous $$\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)$$
That the given expression is undefined and that the expression makes sense only when
$$\lim_\limits{(l,m)\to(\infty,\infty)}\sum_{k=1}^l\prod_{n=k}^m(\frac{9n}{9n+1})$$
exists, in this case it doesn't exist because two different paths give different results.
A: Note that
$$
\exp\left(\frac{-1}{9n+1}\right)\geq1-\frac{1}{9n+1}
$$
for all $n\geq 1$ so that
$$
0=\exp\left(\sum_{n=k}^{\infty}\frac{-1}{9n+1}\right)\geq \prod_{n=k}^{\infty}\left(1-\frac{1}{9n+1}\right)
$$
for each $k\geq 1$. Hence
$$
\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)=0.
$$
