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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?
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    $\begingroup$ @stressed-out But in this case aren't you actually summing zeroes and not infinitesimal numbers? Because $\prod_{n=k}^\infty(\frac{9n}{9n+1})=0$ for every $k$ $\endgroup$ – kingW3 Nov 30 '17 at 17:09
  • $\begingroup$ @kingW3: You are right. :) $\endgroup$ – stressed out Nov 30 '17 at 17:12
  • $\begingroup$ Does the '0 for every k' still count if it is in an infinite sum in which k goes up? $\endgroup$ – Martijn Weterings Nov 30 '17 at 17:13
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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.

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  • $\begingroup$ King, you replicated the second reasoning by adding the variable $a_k$, but could you elaborate some more on it? I have added to the question three points of particular interest. $\endgroup$ – Martijn Weterings Nov 30 '17 at 23:46
  • $\begingroup$ @MartijnWeterings I've edited my answer, hope I answered your questions. $\endgroup$ – kingW3 Dec 1 '17 at 14:43
  • $\begingroup$ thank you very much for your additions. 1) With regard to the first point I will edit my answer (to add my visual interpretation of the problem). I get that we l-l versus l-m is clearly different, however why it still matters when we let both get to the same $\infty$ is unclear. Your final expression $\lim_{l,m \to \infty,\infty}$, and the consideration of a path, is nice in that sense. $\endgroup$ – Martijn Weterings Dec 1 '17 at 14:52
  • $\begingroup$ Your 2) point does not touch entirely what I am wondering about. If we make the proof for every k, then this is only for every finite k right? It feels a bit unsatisfying. That the product term goes to zero is not strange to me. But I wonder about our ability to plug in zero's for all $k$, also when we go to infinity. $\endgroup$ – Martijn Weterings Dec 1 '17 at 14:54
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    $\begingroup$ @MartijnWeterings Sorry for a really late response I was quite busy, I agree that it feels unsatisfying it's mostly because $\prod_{k=\infty}^\infty$ is ambiguous and the most reasonable way is to define $\lim_{n\to\infty}\prod_{k=n}^\infty$ as the limit of partial product where $a_n=\prod_{k=n}^\infty$. $\endgroup$ – kingW3 Dec 8 '17 at 12:49
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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 $$ \sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)=0. $$

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  • $\begingroup$ Why is the LHS of the second inequality equal to $0$? $\endgroup$ – stressed out Nov 30 '17 at 17:06
  • $\begingroup$ I get that the product term $\prod_{n=k}^\infty (\frac{9n}{9n+1})$ is zero for any finite k (should we say exactly zero or almost zero?). But what happens if you make an infinite sum of those product terms while also making the lower subscript going towards infinity? Could you ellaborate on that. $\endgroup$ – Martijn Weterings Nov 30 '17 at 17:09

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