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Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {\sin 2n}{1 + 2^n} $

Convergent

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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Boston College

Okay, So, down in the denominator, we have exponential growth and up top. We just have sign of to end so intuitively, this should be clear that this is going to converge and it looks like it's going to converge fast. Looks like it should converge fast enough that it will actually converge. Absolutely. And sometimes it's easier to prove absolute convergence, and it is to just prove regular convergence. And if you are able to prove absolute convergence than that implies regular convergence as well, that's going to be the approach that we're going to take here. So it starts with just the intuition of, you know, assuming that this is going to converge pretty hard, hard enough, Teo, where just converges, absolutely. And in the observation that absolute convergence in this case is going to be something that is easier to show. It'LL be easier to show because when we take the absolute value here, opposition be equal. We should be getting something that's less than one over one plus to the end, right, because sign can't get any bigger than one. And now this is something that looks much nicer to work with, and we can do Ah, another in equality here replaces one plus two to the end with just to the end. Because remember, if we replace the denominator with something smaller, we're going to get something bigger. That's what's happening here. To the end is smaller than one plus two to Ian. But since it's in the denominator, rain it up with something that's larger and then we know that this is well, this is finite. This is finite and all of these terms of positive So all of these terms are positive and we're bounded. So if all of our terms are positive and we're bounded, then it means that we have convergence, okay? And if we have convergence of this thing, well, that means that we have absolute convergence. And if we have absolute convergence, Madam Clise regular convergence Absolute convergence of this guy gives us regular convergence of this guy