Consider the series ∑_( = 1)^(∞) (−1)^( + 2)/(4 − 1). Determine whether the series is absolutely convergent, conditionally convergent, or divergent. 04:20
Consider the series the sum from equals one to ∞ of negative one to the add two power over four minus one. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Let’s begin with a reminder of what it means for a series to be absolutely convergent or conditionally convergent. A series the sum of is absolutely convergent if the series the sum of the absolute value of is convergent. And it’s conditionally convergent if the series of absolute values diverges but the series itself still converges.
So let’s start by checking whether the series is absolutely convergent or not. This means testing for the convergence of the series the sum from equals one to ∞ of the absolute value of negative one to the add two power over four minus one. One thing to notice is that negative one to the add two power will always give us either one or negative one, depending on the value of .
So taking the absolute value of negative one to the add two power will always give us one. Also, for values of greater than or equal to one, four minus one will always be positive. So we can actually rewrite this series as the sum from equals one to ∞ of one over four minus one. So we need to test this for convergence. We can do this using the comparison test, which tells us that for two series and , both positive sequences with less than or equal to , then if the series converges, the series also converges. But if the series diverges, then the series also diverges.
We could compare our series with the series from equals one to ∞ of one over four . We can see that this is less than or equal to the series the sum from equals one to ∞ of one over four minus one. This works for all positive . If we find that the series the sum from equals one to ∞ of one over four diverges, then we can conclude that our series also diverges. And actually, we can be rewrite the series the sum from equals one to ∞ of one over four by using the constant multiplication rule. This is just the same as one over four multiplied by the series the sum from equals one to ∞ of one over . And this is a series that we recognize. It’s just the harmonic series. And we know this to be divergent. Therefore, the series that we’re looking at is also divergent by the comparison test.
So because the series of absolute values is divergent, the original series in the question is not absolutely convergent. Even though this series is not absolutely convergent, it may still be conditionally convergent. So that’s what we’ll check for next. So I’ll just clear some space to do that. To check for conditional convergence, we check whether the original series is convergent or not. One thing to notice about this series is that the negative one to the add two power creates an alternating effect between positive and negative values. So we can use the alternating series test.
The alternating series tests says that for a series the sum of , where equals negative one to the add one power multiplied by , where is greater than or equal to zero for all , if the limit as approaches ∞ of equals zero and the sequence is a decreasing sequence, then the series is convergent.
For this series, it’s the negative one to the add two power that creates the alternating effect. So our is equal to one over four minus one. We need to check that as approaches ∞, this is going to give us zero and that this is a decreasing sequence. Well, as approaches ∞, four minus one will approach ∞. So the limit as approaches ∞ of one over four minus one is just zero.
We can also see that by looking at the first few terms of this sequence, this is a decreasing sequence. So we can conclude that the series the sum from equals one to ∞ of negative one to the add two power over four minus one is convergent. So we found the series to be convergent but not absolutely convergent. So the series is conditionally convergent.
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So let’s start by checking whether the series is absolutely convergent or not. This means testing for the convergence of the series the sum from equals one to ∞ of the absolute value of negative one to the add two power over four minus one. One thing to notice is that negative one to the add two power will always give us either one or negative one, depending on the value of .
For this series, it’s the negative one to the add two power that creates the alternating effect. So our is equal to one over four minus one. We need to check that as approaches ∞, this is going to give us zero and that this is a decreasing sequence. Well, as approaches ∞, four minus one will approach ∞. So the limit as approaches ∞ of one over four minus one is just zero.
The alternating series tests says that for a series the sum of , where equals negative one to the add one power multiplied by , where is greater than or equal to zero for all , if the limit as approaches ∞ of equals zero and the sequence is a decreasing sequence, then the series is convergent.
So because the series of absolute values is divergent, the original series in the question is not absolutely convergent. Even though this series is not absolutely convergent, it may still be conditionally convergent. So that’s what we’ll check for next. So I’ll just clear some space to do that. To check for conditional convergence, we check whether the original series is convergent or not. One thing to notice about this series is that the negative one to the add two power creates an alternating effect between positive and negative values. So we can use the alternating series test.
We can also see that by looking at the first few terms of this sequence, this is a decreasing sequence. So we can conclude that the series the sum from equals one to ∞ of negative one to the add two power over four minus one is convergent. So we found the series to be convergent but not absolutely convergent. So the series is conditionally convergent.
Said differently, if a series converges, then the series must also converge. It is not hard to see why this is true. The terms of any sequence (possibly containing negative terms) satisfy the inequalities If we assume that converges, then must also converge by the Comparison Test. But then the series converges as well, as it is the difference of a pair of convergent series: Does the series converge?
The basic question we wish to answer about a series is whether or not the series converges. If a series has both positive and negative terms, we can refine this question and ask whether or not the series converges when all terms are replaced by their absolute values. This is the distinction between absolute and conditional convergence, which we explore in this section. Recall that the basic question about a series that we seek to answer is “does the series converge?” It turns out that if the series contains negative terms, there is an interesting refinement of this question. This is illustrated by the following example. We have seen that the alternating harmonic series converges. On the other hand, if we construct a new series by taking the absolute value of each term, we obtain That is, we obtain the standard harmonic series, which is one of our favorite examples of a divergent series.
The series contains both positive and negative terms, but it is not alternating. This makes it difficult to apply our standard tests to determine whether the series converges directly. On the other hand, consider the series By design, all of its terms are nonnegative. Moreover, since , we have the comparison It follows by the Comparison Test that converges. We conclude that converges absolutely, and the Absolute Convergence Theorem implies that it must therefore converge.
If ‘(‘sum |a_n|’) converges we say that ‘(‘sum a_n’) converges absolutely. To say that ‘(‘sum a_n’) converges absolutely is to say that the terms of the series get small (in absolute value) quickly enough to guarantee that the series converges, regardless of whether any of the terms cancel each other. For example ‘(‘ds’sum_{n=1}^’infty (-1)^{n-1} {1’over n^2}’) converges absolutely.
Roughly speaking there are two ways for a series to converge: As in the case of ‘(‘sum 1/n^2’text{,}’) the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of ‘(‘ds ‘sum (-1)^{n-1}/n’text{,}’) the terms dont get small fast enough (‘(‘sum 1/n’) diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. You might guess from what weve seen that if the terms get small fast enough that the sum of their absolute values converges, then the series will still converge regardless of which terms are actually positive or negative. This leads us to the following theorem.
Looking at the leading terms in the numerator and denominator of ‘(a_n’text{,}’) we speculate to compare this series with the harmonic series: ‘begin{equation*} ‘begin{split} ‘text{ Left side } ‘qquad’amp ‘qquad ‘text{ Right side } ” ‘frac{1}{n} ‘qquad’amp ‘qquad ‘frac{3n+4}{2n^2+3n+5} ” 2n^2+3n+5 ‘qquad’amp ‘qquad n(3n+4) ” 2n^2+3n+5 ‘qquad’amp ‘qquad 3n^2+4n ” 5 ‘qquad’amp ‘qquad n^2+n ‘end{split} ‘end{equation*}
FAQ
Which series is absolutely convergent conditionally convergent or divergent?
Is absolutely convergent convergent?
Can a series be both absolutely and conditionally convergent?
How do you determine conditional and absolute convergence?
