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Numerade Educator



Problem 15 Easy Difficulty

Test the series for convergence or divergence.
$ \displaystyle \sum_{n = 0}^{\infty} \frac {\sin(n + \frac {1}{2}) \pi}{1 + \sqrt {n}} $




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Video Transcript

just test the Siri's for convergence or die virgins. So first, let's convince ourselves that this is actually alternating. So the way to see this is just to go ahead and plug in the first few terms. So the first one is an equal zero. You just get pile or two in the sign, one on the bottom and then for the next one signed three Pi over too. Signed five. High Over, too. No, And then let's do one more signed seven Pi over, too. So that's when we plug in three. And then we'LL keep going. In this fashion, we see that the denominator is always positive, but in the numerator we have one and then we have minus one. Then we have one minus one and so on. So, really, the Siri's could have been just ran in simpler terms if they would have just written. It is negative one to the end over one plus radical, and it's the same exact Siri's because the numerator is always wonder minus one, and it starts off being positive one. That's why I have just in here instead of and minus one. So now, in either case, whatever series you want to use? We should call Bien to be one over one plus radical in. And here we just showed it is alternating. So this suggests we can try the all trading Siri's test from this section in the book. So first we define our being to be the positive far So bien is just the absolute value of your Anne. So it doesn't matter which one here you're calling a N If you want to call that a hand or this a end in either case, bien will be one over one plus ruin. So our first condition is that it's positive. So here, just by definition, positive over a positive is positive. The second condition. We need a limit of being to be zero, but this is clear. As an gets Lars, the numerator is only one. Where is the denominator goes to infinity. So we get a zero there and the third condition that we need is that the sequence bien is the racing. So in other words, we need B and plus one less than or equal to B n. That's equivalent to one plus or one over one plus rule and plus form less than or equal to one plus ruin. And this last inequality is true sense. The denominator on left is larger. Any time you have a larger the dominator, your fraction is the hole gets smaller. The new writer stayed the same but then nominated a smaller because we took off the one inside the radical. So this verifies that the sequence being is decreasing, and we've just shown that this bien satisfies all the conditions for the alternating Siri's test. Therefore the Siri's converges and the reason why, by the alternating serious test, and that's our final answer.