= m For example, take the series. On the other hand, if the second series is divergent either because its value is infinite or it doesn’t have a value then adding a finite number onto this will not change that fact and so the original series will be divergent. $\begingroup$ I'm not suggesting a geometric proof for this particular problem be presented in lieu of introducing other topics; instead, I agree with your (apparent) fondness for linking different theories, so I enjoy when proofs can be carried out in multiple ways. 1 Hence the original series is divergent. The point of this problem is really just to acknowledge that it is in fact an alternating series. Both conditions are met and so by the Alternating Series Test the series must converge. First, this is (hopefully) clearly an alternating series with, \[{b_n} = \frac{1}{{7 + 2n}}\] and it should pretty obvious the \({b_n}\) are positive and so we know that we can use the Alternating Series Test on this series. So, \(\left\{ {{s_{2n}}} \right\}\) is an increasing sequence. Since it’s not clear which of these will win out we will need to resort to Calculus I techniques to show that the terms decrease. ≤ , where n for each ; . − 1 Let’s suppose that for \(1 \le n \le N\) \(\left\{ {{b_n}} \right\}\) is not decreasing and that for \(n \ge N + 1\) \(\left\{ {{b_n}} \right\}\) is decreasing. The series can then be written as. n We want to show $\sum_{k=1}^{\infty} (-1)^{k+1}a_k$ converges. Also note that the assumption here is that we have \({a_n} = {\left( { - 1} \right)^{n+1}}{b_n}\). 1 Proof of Alternating Series Test Without loss of generality we can assume that the series starts at \(n = 1\). 2 L a 1 This understanding leads immediately to an error bound of partial sums, shown below. − Now, all that we need to do is run through the two conditions in the test. n Let's say that I have some series, some infinite series. 1 For example, the series (9.5.1) ∑ n = 1 ∞ (− 1 2) n = − 1 2 + 1 4 − 1 8 + 1 16 − … + I get that it’s a part of the proof, but if the series goes to 0 as n tends to infinity, then it has to decrease. + The alternating series test, proved below the next box, is very simple. Remember, that is NOT necessarily true for non-alternating series. ) Finally, in the examples all we really needed was for the \({b_n}\) to be positive and decreasing eventually but for this proof to work we really do need them to be positive and decreasing for all \(n\). | by splitting into two cases. Now, let’s take a look at the even partial sums. lim The first series is a finite sum (no matter how large \(N\) is) of finite terms and so we can compute its value and it will be finite. n ≥ S For instance. We only need to require that the series terms will eventually be decreasing since we can always strip out the first few terms that aren’t actually decreasing and look only at the terms that are actually decreasing. It s pretty Here is the substance of the proof. The two conditions of the test are met and so by the Alternating Series Test the series is convergent. {\displaystyle S_{2m}=\sum _{n=1}^{2m}(-1)^{n-1}a_{n}} a Definition. The Alternating Series Test A series whose terms alternate between positive and negative values is an alternating series. The Alternating Series Test (described below) will prove. k a with odd number of terms, and ) 3rd paragraph: That wasn't really what my concern was. For an alternating series in which the odd terms in the sequence of partial sums are decreasing and bounded below. n The Alternating Series Test This handout presents the solution to Exercise #2.7.1, which asks for a proof of the Alter-nating Series Test using the Cauchy Criterion for series (Theorem 2.7.2). n → Now, the second part of this clearly is going to 1 as \(n \to \infty \) while the first part just alternates between 1 and -1. n The first is outside the bound of our series so we won’t need to worry about that one. Alternating Series test If the alternating series X1 n=1 ( 1)n 1b n = b 1 b 2 + b 3 b 4 + ::: ;b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. Without the \(\pi\) we couldn’t do this and if \(n\) wasn’t guaranteed to be an integer we couldn’t do this. For an alternative proof using Cauchy's convergence test, see Alternating series. then the alternating series converges. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. ∞ This test does not prove absolute convergence. Note that \(x = - 4\) is not a critical point because the function is not defined at \(x = - 4\). n From the proof of the Alternating Series Test we know that $S$ lies between any two consecutive partial sums $s_n$ and $s_{n+1}$. Each of the quantities in parenthesis are positive and by assumption we know that \({b_{2n}}\) is also positive. Increasing \(n\) to \(n + 1\) will increase both the numerator and the denominator. 1 First, unlike the Integral Test and the Comparison/Limit Comparison Test, this test will only tell us when a series converges and not if a series will diverge. So, this tells us that \({s_{2n}} \le {b_1}\) for all \(n\). 0 + Free ebook http://bookboon.com/en/learn-calculus-2-on-your-mobile-device-ebook Example and proof of alternating series test. Alternating series written by btechtuition Prove the alternating series test from the Cauchy convergence theorem for series. 1 If not we could modify the proof below to meet the new starting place or we could do an index shift to get the series to start at \(n = 1\). It is not immediately clear that these terms will decrease. However, it does show us how we can at least convince ourselves that the overall limit does not exist (even if it won’t be a direct proof of that fact). n If a series fails the second requirement for the alternating series test (that the series eventually decreases), that means the series doesn’t converge. Let A n ( 8 n = 1 be a sequence of real numbers that obeys (i) A n ě 0 for all n ě 1 and (ii) A n + 1 ď A n for all n ě 1 (i.e. ∑ It is very important to always check the conditions for a particular series test prior to actually using the test. | All of the conditions in the test, namely convergence to zero and monotonicity, should be met in order for the conclusion to be true. 1 Here is a convergence test for alternating series that exploits this structure, and that is really easy to apply. ( The theorem known as "Leibniz Test" or the alternating series test tells us that an alternating series will converge if the terms an converge to 0 monotonically. − We would like to show Increasing the numerator says the term should also increase while increasing the denominator says that the term should decrease. {\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}\!} n Moreover, let L denote the sum of the series, then the partial sum. Note that, in practice, we don’t actually strip out the terms that aren’t decreasing. 2nd paragraph: Yup, I saw in my textbook, too, that the decreasing part was necessary for the proof. − we see from the graph below that because the values of b n are decreasing, the partial sums of the series cluster about some point in the interval [0;b 1]. a − a ) n ... Then the series converges if . {\displaystyle |a_{n}|} That's what I was saying, that you can't just alter the alternating property to say that it's an alternating series. {\displaystyle \lim _{n\rightarrow \infty }a_{n}=0} n also converges to L. The odd partial sums decrease monotonically: while the even partial sums increase monotonically: both because an decreases monotonically with n. Moreover, since an are positive, So, let’s assume that its limit is \(s\) or. ( k Both conditions are met and so by the Alternating Series Test the series must be converging. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion. | ≥ Now, there are two critical points for this function, \(x = 0\), and \(x = 4\). Let’s do one more example just to make a point. n Give a proof or counterexample. {\displaystyle S_{2n}={\frac {2}{1}}+{\frac {2}{2}}+{\frac {2}{3}}+\cdots +{\frac {2}{n-1}}} (a) Prove the Alternating Series Test by showing that (sn) is a Cauchysequence. Since this condition isn’t met we’ll need to use another test to check convergence. And the “structure” in the partial sum & remainder is: The proof is similar to the proof for the alternating harmonic series. ∑ where either all an are positive or all an are negative, is called an alternating series. However, monotonicity is not present and we cannot apply the test. a In the previous example it was easy to see that the series terms decreased since increasing \(n\) only increased the denominator for the term and hence made the term smaller. m A visual proof of alternating series … 2 So, we now know that both \(\left\{ {{s_{2n}}} \right\}\) and \(\left\{ {{s_{2n + 1}}} \right\}\) are convergent sequences and they both have the same limit and so we also know that \(\left\{ {{s_n}} \right\}\) is a convergent sequence with a limit of \(s\). = ∞ 1 Splitting this limit like this can’t be done because this operation requires that both limits exist and while the second one does the first clearly does not. the series \(\sum {{a_n}} \) is convergent. = For that, we the series may converge or diverge. alternating series button The divergence test will usually be the first test of whether or not a series converges. We describe next a generalization of the comparison test. = S 1 Next, we can quickly determine the limit of the sequence of odd partial sums, \(\left\{ {{s_{2n + 1}}} \right\}\), as follows. We’ll see an example of this in a bit. Let $k,k'\in\mathbb{N}$ and let $a_k$ be a decreasing monotone sequence of non negative numbers such that $a_k\rightarrow 0$. However, it doesn’t necessarily mean that the series diverges.The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i.e., if a1 ≥ a2 ≥ … and if the series converges. 2 Suppose that we have a series \(\sum {{a_n}} \) and either \({a_n} = {\left( { - 1} \right)^n}{b_n}\) or \({a_n} = {\left( { - 1} \right)^{n + 1}}{b_n}\) where \({b_n} \ge 0\) for all \(n\). = 2 follows by taking the negative.)[1]. Prove the following variant of the Alternating Series Test If b n is a positive from CALCULUS 2502 at Western University ( k In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Alternating Series and Leibniz’s Test Let a 1;a 2;a 3;::: be a sequence of positive numbers. + lim we have {\displaystyle S_{2m+1}-S_{2m}=a_{2m+1}\geq 0} Both cases rely essentially on the last inequality derived in the previous proof. the last term is a plus (minus) term, then the partial sum is above (below) the final limit. {\displaystyle \left|S_{k}-L\right|\leq a_{k+1}\!} − | and so we can see that the function in increasing on \(0 \le x \le 4\) and decreasing on \(x \ge 4\). The point of all this is that we don’t need to require that the series terms be decreasing for all \(n\). Do not just make the assumption that the terms will be decreasing and let it go at that. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion. This in turn tells us that \(\sum {{a_n}} \) is convergent. So, the first condition isn’t met and so there is no reason to check the second. − m {\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1}a_{n}\!} All that is required is that eventually we will have \({b_n} \ge {b_{n + 1}}\) for all \(n\) after some point. 2 To get the proof for \({a_n} = {\left( { - 1} \right)^{n}}{b_n}\) we only need to make minor modifications of the proof and so will not give that proof. Is the alternating series test still valid if "strictly decreasing" is omitted? Then if. n Posted on June 18, 2012 by Alex Nelson. In general however, we will need to resort to Calculus I techniques to prove the series terms decrease. There are of course many others, but they all follow the same basic pattern of reducing to one of the first two forms given. It follows that $$ |S-s_n| \leq | s_{n+1} – s_n | \leq a_{n+1} $$ as desired. − I hope that this was helpful. n S = Suppose we are given a series of the form $${\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1}a_{n}\! Proof. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. We will prove that both the partial sums Finally, they must converge to the same number because, Call the limit L, then the monotone convergence theorem also tells us extra information that. The second condition requires some work however. a If you aren’t sure of this you can easily convince yourself that this is correct by plugging in a few values of \(n\) and checking. + Proof integral test, proof alternating series test Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator There are a couple of things to note about this test. k ⋯ Other Alternating Series Tests → Alternating Series Test. n n + and 2 ( Without loss of generality we can assume that the series starts at \(n = 1\). S m n In this case we have. − Actually the series is divergent. − → Edit: rephrase question Explanation of the Alternating Series Test a little bit more concrete. If the second series has a finite value then the sum of two finite values is also finite and so the original series will converge to a finite value. 1 = ∑ It is possible for the first few terms of a series to increase and still have the test be valid. Next, we can also write the general term as. We describe a picture proof of the Alternating Series Test which uses simple comparisons of areas of rectangles to establish convergence. 1 ∞ 1 {\displaystyle \lim _{n\to \infty }a_{n}=0} S . ∑ = 2 a }$$ follows by taking the negative.) Using the test points. Posts about Raabe’s test and Logarithmic test(without proof). A proof of this test is at the end of the section. For a second let’s consider the following. 1 ( Sn is the first n terms, and Rn is from the n+1 term to the rest terms.) Secondly, in the second condition all that we need to require is that the series terms, \({b_n}\) will be eventually decreasing. Proof: Let $\epsilon>0$. + ) 2 approximates L with error bounded by the next omitted term: Suppose we are given a series of the form So, the divergence test requires us to compute the following limit. The proof is similar to the proof for the alternating harmonic series. What happens to ? That won’t change how the test works however so we won’t worry about that. n − − The signs are alternating and the terms tend to zero. 2 A series $\sum a_n$ is called absolutely convergent if the series $\sum |a_n|$ converges. Let’s close this section out with a proof of the Alternating Series Test. 2 The test that we are going to look into in this section will be a test for alternating series. + A series of the form a 1 a 2 + a 3 a 4 + a 5 a 6 + ::: is said to be alternating because of the alternating sign pattern. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. 2 n (c) Consider the subsequences (s2n) and (s2n+1), and show how the Monotone Convergence Theorem leads to a third proof for the Alternating Series Test. 0 = with even number of terms, converge to the same number L. Thus the usual partial sum An alternating series is any series, \(\sum {{a_n}} \), for which the series terms can be written in one of the following two forms. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. 1 Also, the \({\left( { - 1} \right)^{n + 1}}\) could be \({\left( { - 1} \right)^n}\) or any other form of alternating sign and we’d still call it an Alternating Harmonic Series. a The convergence of the series will depend solely on the convergence of the second (infinite) series. For a generalization, see Dirichlet's test. There are many other ways to deal with the alternating sign, but they can all be written as one of the two forms above. S \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\) and, \(\left\{ {{b_n}} \right\}\) is a decreasing sequence. 1 But what about when we let be anything? {\displaystyle S_{2m+1}=\sum _{n=1}^{2m+1}(-1)^{n-1}a_{n}} the sequence is monotone decreasing) and (iii) lim n Ñ8 A n = 0. If you should happen to run into a different form than the first two, don’t worry about converting it to one of those forms, just be aware that it can be and so the test from this section can be used. ) Similarly, the sequence of even partial sum converges too. n It says that if, as n→∞, the terms of an alternating series decrease to zero, then the series converges. + Notice that in this case the exponent on the “-1” isn’t \(n\) or \(n + 1\). Let's say it goes from N equals K to infinity of A sub N. Let's say I can write it as or I can rewrite A sub N. So let's say A sub N, I can write. (The series a 1 + a 2 a 3 + ::: is also alternating, but it is more reassuring to start summation with a positive term.) ∑ n {\displaystyle S_{k}=\sum _{n=1}^{k}(-1)^{n-1}a_{n}} Let $(a_n)$ be a decreasing sequence that converges to $0$. = for any m. This means the partial sums of an alternating series also "alternates" above and below the final limit. 3 Proof… which is twice the partial sum of the harmonic series, which is divergent. Why does the decreasing thing need to be a separate condition? a 0 The series from the previous example is sometimes called the Alternating Harmonic Series. All we do is check that eventually the series terms are decreasing and then apply the test. 1 Alternating Series test If the alternating series X1 n=1 ( n1) 1b n = b 1 b 2 + b 3 b 4 + ::: b n >0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. ∞ for all natural numbers n. (The case n {\displaystyle a_{n}\geq a_{n+1}} 1 2 If not we could modify the proof below to meet the new starting place or we could do an index shift to get the series to start at \(n = 1\). {\displaystyle S_{2n}} 1 After defining alternating series, we introduce the alternating series test to determine whether such a series converges. + First, identify the \({b_n}\) for the test. decreases monotonically[1] and Let’s start with the following function and its derivative. ( n 2 We now know that \(\left\{ {{s_{2n}}} \right\}\) is an increasing sequence that is bounded above and so we know that it must also converge. The alternating series test then says: if So, as \(n \to \infty \) the terms are alternating between positive and negative values that are getting closer and closer to 1 and -1 respectively. https://en.wikipedia.org/w/index.php?title=Alternating_series_test&oldid=996903516, Creative Commons Attribution-ShareAlike License, This page was last edited on 29 December 2020, at 03:30. 2 a Therefore, since \(f\left( n \right) = {b_n}\) we know as well that the \({b_n}\) are also increasing on \(0 \le n \le 4\) and decreasing on \(n \ge 4\). Absolute and Conditional Convergence. First, notice that because the terms of the sequence are decreasing for any two successive terms we can say. I am somewhat stuck on this proof of the alternating series test, could you please point me to the right direction ? 1 = m 727–730. S The \({b_n}\) are then eventually decreasing and so the second condition is met. m }$$, where $${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0}$$ and $${\displaystyle a_{n}\geq a_{n+1}}$$ for all natural numbers n. (The case $${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}\! Proof: Suppose the sequence converges to zero and is monotone decreasing. We have discussed whether a series will converge, when for each . a (b) Supply another proof for this result using the Nested Interval Property(Theorem 1.4.1). The proof follows the idea given by James Stewart (2012) “Calculus: Early Transcendentals, Seventh Edition” pp. More precisely, when there is an odd (even) number of terms, i.e. n In these cases where the first condition isn’t met it is usually best to use the divergence test. = Alternating series test: If $\\{a_n\\}$ is positive and strictly decreasing, and $\\lim a_n=0$, then $\\sum(-1)^n a_n$ converges. In most cases, an alternation series #sum_{n=0}^infty(-1)^nb_n# fails Alternating Series Test by violating #lim_{n to infty}b_n=0#.If that is the case, you may conclude that the series diverges by Divergence (Nth Term) Test. As the previous example has shown, we sometimes need to do a fair amount of work to show that the terms are decreasing. m 1 Indeed, for the partial sum It is important that the series truly alternates, that is each positive term is followed by a negative one, and visa versa. To see why this is consider the following series. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. proof of alternating series test The series has partial sum S2⁢n+2=a1-a2+a3-+…-a2⁢n+a2⁢n+1-a2⁢n+2, where the aj’s are all nonnegative and nonincreasing. 1 In order for limits to exist we know that the terms need to settle down to a single number and since these clearly don’t this limit doesn’t exist and so by the Divergence Test this series diverges. S We’re learning alternating series test and the whole class is confused on why the series needs to be decreasing to pass the test. To see this we need to acknowledge that. The infinite series in question is: a 1 −a 2 +a This limit can be somewhat tricky to evaluate. It should be pointed out that the rewrite we did in previous example only works because \(n\) is an integer and because of the presence of the \(\pi\). Thus we can collect these facts to form the following suggestive inequality: Now, note that a1 − a2 is a lower bound of the monotonically decreasing sequence S2m+1, the monotone convergence theorem then implies that this sequence converges as m approaches infinity. Prove that the series $$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$$ converges by showing that the sequnce of partial sums is a cauchy sequence. Figure \(\PageIndex{2}\): For an alternating series \( b_1−b_2+b_3−⋯\) in which \( b_1>b_2>b_3>⋯\), the odd terms \( S_{2k+1}\) in the sequence of partial sums are decreasing and bounded below.

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