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I have a 2-indexed sequence $F^n_m$ where $n,m$ are natural numbers and I am concerned about the behavior as $n\to\infty$ and/or $m\to\infty$. The sequence is expressed as

$$F^n_m=\prod_{k=1}^na_{mk}$$

For each finite $n$ and $k$, as $m \to \infty, a_{mk}$ converges to a finite nonzero limit, say $a_{\infty k}$ and therefore

$$F^n_\infty=\lim_{m \to \infty}F_m^n=\prod_{k=1}^na_{\infty k}$$

exists and is finite as a finite product of converging sequences. Now if we let $n \to \infty$ I have shown that

$$F=\lim_{n \to \infty} F_{\infty}^{n}$$

exists and is expressible as the corresponding infinite product (which is by the way the product representation of Riemann's zeta function). My question is what happens when we approach the limit by letting $n \to \infty$ first and then $m \to \infty$. In fact I haven't found a closed form for

$$F_m^\infty=\lim_{n \to \infty}F_m^n=\prod_{k=1}^{\infty}a_{mk}$$

and I don't know if it even converges. Are there any conditions under which we can do that, then take the limit as $m \to \infty$ and come up with the same answer?

And what about if we approach $\infty$ "diagonally"? By that I mean given a strictly increasing sequence $p(m)$ of integers and consider

$$\lim_{m \to \infty} F_m^{p(m)}$$

Are we going to get a value that depends on $p$ or the same value for all such sequences? And under what conditions? There is a theorem that states that the infinite product

$$\prod_{k=1}^\infty(1+b_k)$$

converges or diverges according to the sum

$$\sum_{k=1}^\infty b_k$$

but it didn't quite do it...

I am looking for rigorous as well as not-so-rigorous approach, anything really!

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1 Answer 1

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Looks like an iterated limit, right?
There has not been given an accepted (i.e. satisfactory) answer to the following question:

Please read the question, the answer and the comments in there and make up your mind. Anyway, it is not a standard result in common mathematics - don't ask me why.
In your case the Commutativity of iterated limits Theorem (see above reference) says: $$ \lim_{n \to \infty} \left[ \lim_{m \to \infty} F^n_m=\prod_{k=1}^na_{mk} \right] = \lim_{m \to \infty} \left[ \lim_{n \to \infty} F^n_m=\prod_{k=1}^na_{mk} \right] $$

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