 The formal definition of a limit of a sequence has one major drawback. You have to know the limit before you can apply it. This will be an even bigger problem with infinite series. To avoid this, let's consider a few things. So I suppose our sequence has limit l. Then, by the definition for any epsilon greater than 0, there is an n, where for all n greater than n, a n minus l absolute value is strictly less than epsilon. Now, if we take m even larger, then it's also true that a m minus l will be less than epsilon. So the sum will be less than 2 epsilon. And remember, the absolute value of a sum is less than or equal to the sum of the absolute values, but that also works the other way, and so we can condense this and make it smaller. And so we get, and we still have the limit l. So this approach doesn't seem to be useful. So let's ignore it for a moment, but keep in mind the fundamental idea of combining the absolute values. So what else could we do? We might remember that the absolute value of a minus b is the same as the absolute value of b minus a. So that means we can rewrite our second absolute value, combine them, and get. Now it's also useful to remember that if you're off by a constant factor, we can just scale it. So instead of being less than epsilon, we can be less than epsilon halves, and we can adjust the rest of the computations accordingly, and find our differences strictly less than epsilon. And this leads to the Cauchy convergence criterion. The sequence converges if the difference between terms can be smaller than epsilon. Now the proceeding actually proves that if a sequence is convergent, it is Cauchy convergent. We can also prove the converse, but you should do that. So let's prove that this sequence converges. So let epsilon greater than 0 be given. For m and n greater than, well I don't know what, we'll figure that out as we go. We want the difference to be less than epsilon. Now we know what our destination is, so let's fill in the roadmap. And as much as I despise bumper sticker mathematics, here's a useful one. If in doubt, factor out. Here we see both terms are multiplied by a power of one third, and we can factor out a power, although we'd rather not have to deal with negative exponents. So remember, you can assume anything you want as long as you make it explicit. We'll assume that m is the larger of the two numbers. And so we can factor this as and now because this is the absolute value of a difference, we can split it. And because these are sine multiplied by something, we know that these are going to be both less than one. So this is going to be less than. And remember, we want this to be less than epsilon. Well now we have an equation that we can solve and we find. And so we know that n and our lower bound has to be log 2 minus log epsilon divided by log 3. And since we can make the difference smaller than epsilon, this proves the sequence converges. One final note, the algebra here is not particularly difficult, so we often omit the details of the algebra and we replace this explicit computation of our lower bound with a phrase like, for sufficiently large n,