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Question Video: Finding the Rate of Change in the Area of a Shrinking Circular Disc Using Related Rates Mathematics • Higher Education

State whether the series β_(π =1)^(β) (β1)^(π + 1) 2/β(π + 1) converges absolutely, conditionally, or not at all.

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Video Transcript

State whether the series the sum from π equals one to β of negative one to the power of π add one multiplied by two over the square root of π add one converges absolutely, conditionally, or not at all.

Firstly, recall that for a series π π, this is absolutely convergent if the series of absolute values converges. And itβs conditionally convergent if the series of absolute values diverges. But the series itself converges. So letβs first of all find out whether this series is absolutely convergent or not. This means testing whether the series from π equals one to β of the absolute value of negative one to the power of π add one multiplied by two over the square root of π add one is convergent or divergent.

Well, negative one to the power of π add one is always going to be one or negative one. But if we take the absolute value, it will always be one. Whereas two over the square root of π add one will always be positive because π runs through positive values. So we can write this as the sum from π equals one to β of two over the square root of π add one. Then, we can use the constant multiplication rule to bring the two to the front of the sum. From here, we need to work out whether this series converges or diverges. One way we can actually do this is with a direct comparison with the harmonic series.

Because for π is greater than two, we have that one over π is less than one over the square root of π add one. And we know that if we have π π less than π π where π π diverges, then π π also diverges. And we know that the sum from π equals one to β of one over π is the harmonic series which diverges. Then, the sum from π equals one to β of one over the square root of π add one also diverges. So we found that the series of absolute values diverges, which means that this series is not absolutely convergent. But it could still be conditionally convergent. So weβre going to test the series itself for convergence. So letβs clear some space.

We can firstly bring the constant two to the front of the sum. And then, if we look at this negative one to the power of π add one, this creates an alternating series because it makes the terms alternate between positive and negative. So we can decide whether this series is convergent or divergent using the alternating series test. Remember that this says for a series π π, where π π is equal to negative one to the power of π add one multiplied by π π, if π π is decreasing and the limit as π approaches β of π π is equal to zero, then the series π π is convergent. So for our series π π is equal to one over the square root of π add one. But is this decreasing? Well, for π π to be decreasing, we need π π to be greater than π π add one.

Well, we can see that as π increases by one, the square root of π add one is going to get bigger. So one over the square root of π add one is going to decrease. So π π is decreasing. Then, we need to check whether the limit as π approaches β of π π is equal to zero. In other words, is the limit as π approaches β of one over the square root of π add one zero? Well, the square root of π add one is increasing as π gets bigger. So this will be one over β. So as π approaches β, then one over the square root of π add one approaches one over β. And so the limit as π approaches β is zero. So both of these conditions are satisfied. So this series is convergent.

Remember that we said that a series is conditionally convergent if the series of absolute values diverges but the series converges. And thatβs exactly what weβve found here. The series of absolute values was divergent. But we found the series itself to be convergent. So we can conclude that this series converges conditionally.

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