This is not a homework problem. I'm doing it for fun. But it is the kind that might appear on homework. 1. The problem statement, all variables and given/known data I'm trying to prove that if lim n→∞ |an+1/an| = L < 1, then [itex]\Sigma[/itex] an converges absolutely and therefore converges. 2. Relevant equations 3. The attempt at a solution Here's my thinking. I feel like I'm on the right track, but I may need some help formalizing my expression of what's happening "in the limit" as n→∞. To show that ∑an converges absolutely, I need to show that ∑|an| converges. My strategy is to show that, because of the condition above, as n→∞ this "tends toward" a geometric series with common ratio < 1 and therefore converges. I have ∑an = |a1| + |a2| + |a3| + ... + |an| = |a1| + |a1||a2/a1| + |a1||a2/a1||a3/a2| + ... + |a1||a2/a1||a3/a2||a4/a3|...|an/an-1| Supposing lim n→∞ |an+1/an| = L < 1, this means that there is some integer N such that for n > N, all the ratios |an+1/an| are equal in the sense that the difference between any two of them can be made arbitrarily small by choosing N appropriately. Therefore I conclude that for n > N, the series above can be written ∑|an| = |a1|(1 + L + L2 + L3 + L4 + ... + Ln) i.e. it is a geometric series (or eventually becomes one beyond N) with common ratio < 1 and therefore converges. And finally, since ∑|an| converges, ∑an converges absolutely and therefore it also converges. Is my reasoning solid here? I am particularly concerned about the conclusion that all the ratios in the series (the ones that multiply |a1| in each term) eventually equal L. I think my intuition here is correct, but I am not well enough attuned to the subtleties of analysis to be confident that my chain of reasoning is unimpeachable. NOTE: I know there must be better, shorter, more elegant, less cumbersome ways of proving this. I'm not interested in those until I have developed my own proof to the utmost. That way I will learn the most from the process. Thanks!