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Numerade Educator



Problem 45 Hard Difficulty

If $ \sum a_n $ is a convergent series with positive terms, is it true that $ \sum \sin(a_n) $ is also convergent?


Yes, it is true.


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Video Transcript

given that this Siri's converges and and positive. Now we'd like to see if the serious convergence since the limit of an well, actually, let me take a step back here. So since converges, this limit has to be zero. This's by the test for diversions. Come on. So there exists and end such that if little and his bankers and begin than zero Weston am less than one. So now, with use the fact that okay, sign X is always less than or equal to X. This is if X is positive. So this implies Signe and less than or equal to an and therefore if little and is bigger than capital and then sign of an It's just some number between zero and one, as long as the input, the angles between zero and one so we can write this sum. It doesn't matter what the starting point is like his aides for more and equals one. No, we can write this now on the second. Siri's over here. We know that this Siri's will converge because this is less than or equal to just an and then we have just a finite sum here. So when we go on to the next page. So this is the whole right hand side is less than or equals who this converges. This is just a finite sum. So it emerges and therefore the Siri's converges by the comparison test. So let me make a remark here, although it is possible that now all of these guys who are bigger than zero what we showed was that by writing it in this form that eventually it is bigger than zero. And that's what really matters here. This is why we split it into two sums and then on the second some here and is bigger than capital in. And this is from what I showed in the previous page. I know that sign was positive for these values of them, and so, really, I used the comparison test on this. Siri's only So this is a finite Siri's and that this over here is just a finite sum. And if I add these two together, the result is that it convergence by the computer success, and that's the final answer