 Determine the convergence of the given series, the sum n equals 1 to infinity of n minus 1 over 2n plus 1. So take this series for a moment, pause this video if necessary, and consider which convergence test might you use on this series to determine whether it's convergent or divergent. Now for me personally, what I'm thinking is something like the following. I have a rational expression here for the sequence n minus 1 over 2n minus 1. If I focus, if I focus just on the leading terms, I have this n over 2n, n over 2n, this sequence is approximately the same thing to that which looks like 1 half. So a comparison of some kind might be useful. Because of the plus 1 on the bottom and the minus 1 on the top, a direct comparison test would be very difficult. A limit comparison test is also possible, but even better yet, when you look at this sequence, this is supposed to be approximately equal to 1 half. Actually, the test of divergence comes to my mind right here. Consider the sequence n minus 1 over 2n plus 1. If we take the limit as n goes to infinity, again, because this is a balanced rational function, this thing will convert, this will be the same limit as n over 2n. You could also use L'Hopital's rule here, but that's kind of overkill. This is going to become just 1 half. In terms of thinking of the comparison test, I actually then identify that, hey, this thing is not going to converge to 0. The limit of the sequence is not 0, and therefore this series is going to be divergent. It's divergent. And what are we using here? We're using the divergence test. And so I think that's perhaps the simplest way to determine the divergence of this series here, but using the divergence test. But as we mentioned before, you might be wanting to use a limit comparison test because at the beginning, it kind of looked like that. And the thought process that got me thinking about the limit comparison test ultimately led to me thinking it was divergence by the divergence test.