# splitting one limit into two?

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.

-
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. –  Yemon Choi Apr 8 '13 at 4:12
Actually, I now realize that the latter expression is also an ok starting point, which is what I needed. Thanks for the help. –  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.
Also, something should be said about denominators staying reasonably away from $0$; well--something :-)