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Summary of Series of Constants an Infinite Series Is a Summation Summary of Series of Constants ∞ ∞ X X An infinite series is a summation an = ai = a1 + a2 + a3 + a4 + ... Note that the variable used for the n=1 i=1 index of summation is irrelevant. The ai’s are the terms of the series. We want to make sense of the sum of this infinite collection of terms. Essentially, we define the sum, S, to be the number that gets approached as one adds more and more terms. To make this precise, we define the partial sums, Pn SN , of the series by letting SN equal the (finite) sum of the first N terms of the series, i.e., SN = i=1 ai = a1 + a2 + a3 + ··· + aN . Then, the sum of the original infinite series is the limit of SN as N approaches infinity, i.e., S = lim SN ; we say N→∞ that the infinite series converges if this limit exists, and diverges if this limit does not exist. If the series converges, then we say that it converges to the sum S. Note that, if one changes/adds/deletes some finite number of terms in a series, then that does not affect whether the series converges or diverges (it does, however, affect the number that one gets for the sum). In particular, in the tests below, all of the series begin at n = 1, but the conclusions about whether a series converges or diverges are all the same if one begins with n = 57 or n = 5, 000, 000, 000, 000, but – remember – leaving off some terms does affect the actual number that a series converges to. Geometric series ∞ X If a and r are fixed real numbers, and a, r 6= 0, then the series ari−1 = a + ar + ar2 + ar3 + ... is called a i=1 geometric series. The initial term is a, and r is the ratio of successive terms, i.e., to obtain the next term in the series, one always multiplies the previous term by the same constant r. It is easy to obtain a formula for the partial sums of a geometric series; one finds N X 1 − rN S = ari−1 = a + ar + ar2 + ··· + arN−1 = a , if r 6= 1, N 1 − r i=1 and SN = Na, if r = 1. It follows that the infinite geometric series converges if and only if −1 < r < 1, i.e., |r| < 1, and when the a geometric series converges, it converges to . 1 − r Telescoping series ∞ ∞ X X √ √ A series of the form f(n) − f(n + 1), for example n − n + 1, is called telescoping. It is easy n=1 n=1 to obtain a formula for SN for telescoping series because all of the middle terms cancel out. We obtain that SN = f(1) − f(N + 1) and, thus, the series converges if and only if lim f(N + 1) exists. N→∞ The n-th term test for divergence ∞ X If lim an 6= 0 (including if lim an does not exist), then an diverges. Note that it is, in general, wrong to n→∞ n→∞ n=1 conclude that a series converges simply because lim an = 0; see the harmonic series below. n→∞ The harmonic series ∞ X 1 1 1 1 The series = 1 + + + + ... is called the harmonic series. This series diverges, despite the fact that n 2 3 4 n=1 lim an = 0. n→∞ Non-negative series ∞ ∞ X X If all the terms an in the series an are non-negative (i.e., an > 0, for all n), then an converges if and only if n=1 n=1 the partial sums of the series are bounded above. In other words, the only way for a series with non-negative terms to diverge is to diverge to +∞. The integral test If f(x) is a continuous, decreasing, positive function and, for every integer n > 1, an = f(n), then the series ∞ X Z ∞ an converges if and only if the improper integral f(x) dx converges. The same result remains true if we n=1 1 ∞ X begin the summation and the integral at the same positive integer other than 1. Note that, even if both an and n=1 Z ∞ f(x) dx converge, they will usually converge to different numbers. 1 p-series ∞ X 1 An important conclusion derived from the integral test is that, if one fixes a number p, then the p-series = np n=1 1 1 1 1 + + + + ... converges if and only if p > 1. 2p 3p 4p The direct comparison test ∞ X Suppose that we have two sets of terms, an’s and bn’s, and that, for all n, 0 6 an 6 bn. Then, if bn converges, n=1 ∞ ∞ ∞ X X X then so does an; in addition, if an diverges, then so does bn. n=1 n=1 n=1 The limit comparison test Suppose that we have two sets of terms, an’s and bn’s, and that, for all n, 0 < an and 0 < bn. Suppose that a lim n = c, where c might be ∞. n→∞ bn ∞ ∞ X X a) If 0 < c < ∞, then the series an and bn do the same thing, i.e., either both series converge or both series n=1 n=1 diverge. ∞ X b) If c = ∞, then, for large n, the an’s must be much larger than the bn’s. Hence, if c = ∞ and bn diverges, then n=1 ∞ X so does an. n=1 ∞ X c) If c = 0, then, for large n, the an’s must be much smaller than the bn’s. Hence, if c = 0 and bn converges, n=1 ∞ X then so does an. n=1 The ratio test ∞ X an+1 Suppose we have a series an, and suppose that the limit L = lim exists or is infinite. Then, if L < 1, n→∞ a n=1 n the series converges. If L > 1 (including if L = ∞), then the series diverges. If L = 1, the test is inconclusive: the series might be convergent or divergent. The root test ∞ X pn Suppose we have a series an, and suppose that the limit L = lim |an| exists or is infinite. Then, if L < 1, n→∞ n=1 the series converges. If L > 1 (including if L = ∞), then the series diverges. If L = 1, the test is inconclusive: the series might be convergent or divergent. Absolute and conditional convergence ∞ ∞ ∞ X X X If |an| converges, then the series an converges, and we say that an converges absolutely. n=1 n=1 n=1 ∞ ∞ X X If the series an converges, but does not converge absolutely, then we say that an converges conditionally. n=1 n=1 ∞ ∞ X X Note that to show conditional convergence, one has to show two things: that an converges and that |an| n=1 n=1 diverges. If one rearranges the terms in a series which converges absolutely, it does not affect the sum. However, by rearranging the terms in a conditionally convergent series, one can make the series converge to any number or even diverge. Alternating series n−1 n If bn > 0 for all n, then the series with terms an = (−1) bn or an = (−1) bn have terms that are alternately positive then negative (or vice-versa); such series are called alternating series. Thus, alternating series are of the form ∞ ∞ X n−1 X n (−1) bn or (−1) bn , where bn > 0. n=1 n=1 Note that bn is the absolute value of the n-th term, i.e., in either case, bn = |an|. For an alternating series, if the bn’s are decreasing and lim bn = 0, then the series converges; in addition, in this case, if the alternating series converges n→∞ to S, then |S − SN | < bN+1, where SN denotes the N-th partial sum of the alternating series. This inequality allows us to estimate the sum of the series, S, by using the partial sums: for all N, SN − bN+1 < S < SN + bN+1..
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