CHAPTER XIX. SERIES. 365. A series is a succession of terms each of which is derived from one or more of the preceding ones by a fixed law. This law is called the law of the series. The number of terms of the series is generally unlimited. Arithmetical and geometrical progressions afford examples of series. 366. A converging series is one in which the sum of the first n terms can not numerically exceed some finite quantity, however great n may be. Thus, 1, 1, 1, 1, 1, etc., is a converging series. 367. A diverging series is one in which n can be taken so large that the sum of the first n terms is numerically greater than any finite quantity. Thus, 1, 2, 3, 4, 5, 6, etc., is a diverging series. 368. When a certain number of terms are given, and the law of the series is known, we may find any term of the series, or the sum of any number of terms. This may generally be done by the method of differences. 369. To find the several orders of differences for any series: Subtract the first term from the second, the second from the third, the third from the fourth, etc.; we shall thus form a new series, which is called the first order of differences. Subtract the first term of this new series from the second, the second from the third, etc.; we shall thus form a third series, called the second order of differences. Proceed in like manner for the third, fourth, etc., orders of differences, and so on till they terminate, or are carried as far as may be thought necessary. Ex. 1. Find the several orders of differences of the series of square numbers 1, 4, 9, 16, etc. Ex. 2. Find the several orders of differences of the series of cube numbers 1, 8, 27, etc. Ex. 3. Find the several orders of differences of the series of fourth powers 1, 16, 81, 256, 625, 1296, etc. Ex. 4. Find the several orders of differences of the series of fifth powers 1, 32, 243, 1024, 3125, 7776, 16807, etc. Ex. 5. Find the several orders of differences of the series of numbers 1, 3, 6, 10, 15, 21, etc. 370. To find the nth term of any series: Let a, b, c, d, e, etc., represent the proposed series. If we subtract each term from the next succeeding one, we shall obtain the first order of differences; if we subtract each term of this new series from the succeeding term, we shall obtain the second order of differences, and so on, as exhibited in the fol lowing table: Series. 1st Diff. 2d Differences. 3d Order of Differences. 4th Order of Differences. Let D', D", D'", D'""', etc., represent the first terms of the several orders of differences. D'=b-a, Then we shall have D"-c-2b+a, D""-d-3c+3b-a, D"""-e-4d+6c-4b+a, " etc., d=a+3D'+3D"+D"". e=a+4D'+6D''+4D'''+D''"', etc. The coefficients of the value of c, the third term of the proposed series, are 1, 2, 1, which are the coefficients of the second power of a binomial; the coefficients of the value of d, the fourth term, are 1, 3, 3, 1, which are the coefficients of the third power of a binomial, and so on. Hence we infer that the coefficients of the nth term of the series are the coefficients of the (n-1)th power of a binomial. If we denote the nth term of the series by Tn, we shall have (n−1) (n—2)Ɗ" + (n-1) (n—2) (n—3)Ɗ'” (n−1)(n−2)D''+ Tn=a+(n−1)D+ 2 +, etc. 2.3 Ex. 1. Find the 12th term of the series 2, 6, 12, 20, 30, etc. The first order of differences, แ 4, 6, 8, 10, etc. second order of differences, 2, 2, 2, etc: "C third order of differences, Here D'=4, D"=2, and D""=0. Hence T12=2+11D'+55D"=2+44+110=156, Ans. 1, 3, 6, 10, 15, 21, etc. Here D'=2, D"=1, a=1, and n=20. Therefore T20=1+19D'+171D"=1+38+171=210, Ans. Ex. 3. Find the thirteenth term of the series 1, 5, 14, 30, 55, 91, etc. Ex. 4. Find the fifteenth term of the series 1, 4, 9, 16, 25, 36, etc. Ex. 5. Find the twentieth term of the series 1, 8, 27, 64, 125, etc. Ex. 6. Find the nth term of the series 1, 3, 6, 10, 15, 21, etc. Ans. n(n+1) Ex. 7. Find the nth term of the series 1, 4, 10, 20, 35, etc. Ex. 8. Find the nth term of the series 1, 5, 15, 35, 70, 126, etc. 371. To find the sum of n terms of any series: Let us assume the series 24 0, a, a+b, a+b+c, a+b+c+d, etc. (1.) Subtracting each term from the next succeeding, we obtain the first order of differences, Now it is clear that the sum of n terms of the series (2) is equal to the (n+1)th term of series (1); and the nth order of differences in series (2) is the (n+1)th order in series (1). If, then, we denote the sum of n terms of series (2) by S, which is the same as the (n+1)th term of (1), we may obtain the value of S from the formula of the preceding article by substituting 0 for a, n+1 for n, Hence a for D', D' for D", etc. S=na+”(n−1)D'+”(n−1)(n−2)D′+n(n−1)(n−2)(n−3)D'', etc. 2.3.4 When any one of the successive orders of differences becomes zero, this formula gives the exact sum of the terms. When no order of differences becomes zero, the formula may still give approximate results, which will, in general, be nearer the truth the greater the number of terms employed. EXAMPLES. 1. Find the sum of 15 terms of the series 1, 3, 6, 10, 15, 21, etc. Here a=1, D'=2, D"=1, D""=0. Therefore S=15+15.14+5.7.13-680, Ans. 2. Find the sum of 20 terms of the series 1, 4, 10, 20, 35, etc. 3. Find the sum of n terms of the series 1, 2, 3, 4, 5, 6, etc. 4. Find the sum of n terms of the series 12, 22, 32, 42, 52, etc. 7. Find the sum of n terms of the series 1.2, 2.3, 3.4, 4.5, 5.6, etc. 372. Interpolation is the process by which, when we have given a certain number of terms of a series, we compute intermediate terms which conform to the law of the series. Interpolation may, in most cases, be effected by the use of the formula of Art. 370. If in this formula we substitute n+1 for n, we shall have Tn+1=a+nD'+ n(n-1) D" + 2 n (n−1)(n−2) D'''+, etc., 2.3 which expresses the value of that term of the series which has n terms before it. When n is a fraction less than unity, Tn+1 |