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Determine whether the series is absolutely conver…

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Problem 2 Easy Difficulty

Determine whether the series is absolutely convergent or conditionally convergent.

$ \displaystyle \sum_ {n = 1}^{\infty} \frac {( - 1)^{n-1)}}{\sqrt{n}} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 6

Absolute Convergence and the Ratio and Root Tests

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Video Transcript

let's determine whether the Siri's this absolutely conversion. So to do that, instead of looking at the Siri's the way that it's written well, go ahead and replace the terms that were adding with their absolute value. So in this case, when we take the absolute value of negative one to the end minus one, this is just one over square then, so that will go here. Now we can rewrite this as one over into the one half, and we see that this is diversion due to the Peters with P equals one half, which is less than one. Anything that's less than equal to one and the P series. It's always diversion. You need to be larger than one, but one half is not larger than one. So we have Mrs Khama so not absolutely commercial. However, we can show that the original Siri's does converge, just not absolutely so to do that. Let's look at this is alternating Siri's. So let's just look at the positive part. The B end equals one over square. Then Now let's apply the alternating Siri's test. There's some conditions that need to be satisfied there. The first one is that you're being is not negative. This is always true for any end. One over the square of event is not negative because it's positive, divided by a positive. The second condition is we need at the limit of the being ghost zero as n goes to infinity, and clearly we can see that that's the case here. As we take end to infinity, the numerator is just one, but the denominator goes to infinity and won over infinity zero. So that's true. And for the third, when I go on to the next page here, you need that bien is decreasing. So eventually this condition has to be true. In our problem, this becomes one over the squared of n plus one less than equal, one over square root of n, and this is always true. If it's not obvious why it's true, you could cross multiply and now you see that that's true. If that's still not obvious to you, that that's truly good square both sides, and now we can see that that's true because that's just a global into zero, less than or equal to one. Therefore, the Siri's, the original Siri's one conversions by the alternative theories test. Now let's make our final conclusion on the next page, since the Siri's Oh, the original series converges, but not absolutely. Our conclusion is that the Siri's is conditionally conversion, and that's our final answer.

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