Suppose I have the limit

$\lim_{m\rightarrow \infty}\frac{\sum_{k=0}^ma_{k,m}}{\sum_{k=0}^mb_{k,m}}$.

When can I write this as

$\lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty} \frac{\sum_{k=0}^ma_{k,n}}{\sum_{k=0}^mb_{k,n}}$?

To be specific, both sums converge to exponentials, which tend to zero as $n\rightarrow$. I'd like to take their ratio before letting $n$ tend to infinity.

  • 1
    $\begingroup$ As stated, I think I can cook up counterexamples, so it might be helpful to be even more specific about what you actually want to prove. $\endgroup$ – Yemon Choi Apr 8 '13 at 4:12
  • $\begingroup$ Actually, I now realize that the latter expression is also an ok starting point, which is what I needed. Thanks for the help. $\endgroup$ – user32851 Apr 8 '13 at 16:22

The first (single) limit is totally blind to terms   $a_{k\ n}\ \ b_{k\ n}$   for all   $k > n$,   while the second (double) limit depends on them. Thus the two limits are hardly related at all.

In other words, the single limit considers finite segments, and the double limit the infinite segments. To compare these two there should be perhaps a relation given between the finite and infinite sums.

Also, something should be said about denominators staying reasonably away from $0$; well--something :-)

(I am surprised that the m-th sums in the single limit case have exactly m terms--I'd expect a more flexible situation).


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