CHAPTER XXIX. SUMMATION OF SERIES. 380. Examples of summation of certain series have occurred in previous chapters; it will be convenient here to give a synopsis of the methods of summation which have already been explained. (i) Arithmetical Progression, Chap. IV. (ii) Geometrical Progression, Chap. V. (iii) Series which are partly arithmetical and partly geometrical, Art. 60. (iv) Sums of the powers of the Natural Numbers and allied Series, Arts. 68 to 75. (v) Summation by means of Undetermined Coefficients, Art. 312. (vi) Recurring Series, Chap. XXIV. We now proceed to discuss methods of greater generality; but in the course of the present chapter it will be seen that some of the foregoing methods may still be usefully employed. 381. If the 7th term of a series can be expressed as the difference of two quantities one of which is the same function of r that the other is of r-1, the sum of the series may be readily found. r and its sum by S, and suppose that any term u can be put in the form v,-,,; then S1 = (v,−v ̧)+(v ̧−v ̧)+(v ̧−v2) +...+(v,−1 − Vμ−2) + (V„ − Vn−1) 2 382. (1 + x) (1 + n + 1.x) Sometimes a suitable transformation may be obtained by separating u into partial fractions by the methods explained in Chap. XXIIÏ. Example. Find the sum of 1 + (1+x) (1+ax) * (1+ax) (1+a*x) + (1 + a2x) (1 + a3x) (1+an−1x) (1+a2x) ̄ ̄ 1+aa ̄1x 1+a"x suppose; By putting 1+a”−1x, 1+a^x equal to zero in succession, we obtain 383. To find the sum of n terms of a series each term of which is composed of r factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression. Let the series be denoted by u1 + U2 + U3 + where un = ...... + Un, (a + nb) (a + n + 1 . b) (a + n + 2 . b) ... (a + n + r − 1.b). Replacing n by -1, we have n un-1 = (a + n − 1 . b) (a + nb) (a + n + 1. b) ... (a+n+r-2.b); .. (a + n − 1 . b) u2 = (a + n + r − 1. b) un-1 = vn, say. Replacing n by n + 1 we have where C is a quantity independent of n, which may be found by ascribing to n some particular value. The above result gives us the following convenient rule: Write down the nth term, affix the next factor at the end, divide by the number of factors thus increased and by the common difference, and add a constant. (+1) b α (r + 1) b u1; it is however better not to quote this result, but to obtain C as above indicated, Example. Find the sum of n terms of the series 1.3.5 +3.5.7+5.7.9+...... The nth term is (2n − 1) (2n + 1) (2n+3); hence by the rule To determine C, put n=1; then the series reduces to its first term, and we have 15= 1.3.5.7 15 +C; whence C=; 8 384. =n (2n3 +8n2+7n - 2), after reduction. The sum of the series in the preceding article may also be found either by the method of Undetermined Coefficients [Art. 312] or in the following manner. -- We have u„ = (2n − 1) (2n + 1) (2n + 3) = 8n3 + 12n3 — 2n − 3 ; ... S1 = 8Σn3 + 12Σn2 – 2Σn – 3n, using the notation of Art. 70; ... S1 = 2n2 (n + 1)2 + 2n (n + 1) (2n + 1) − n (n + 1) −3n 385. It should be noticed that the rule given in Art. 383 is only applicable to cases in which the factors of each term form an arithmetical progression, and the first factors of the several terms are in the same arithmetical progression. Thus the sum of the series 1.3.5 +2.4.6+ 3.5.7 + ...... to n terms, may be found by either of the methods suggested in the preceding article, but not directly by the rule of Art. 383. Here u1 = n (n + 2) (n + 4) · = n (n + 1 + 1 ) (n + 2 + 2) = n (n + 1) (n + 2) + 2n (n + 1) + n (n + 2) + 2n The rule can now be applied to each term ; thus S1 = 1n (n + 1) (n + 2) (n + 3) + n (n + 1) (n + 2) + 3 n (n + 1) + C' = 1 n (n + 1) (n + 4) (n + 5), the constant being zero. 386. To find the sum of n terms of a series each term of which is composed of the reciprocal of the product of r factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression. Let the series be denoted by u1 + U2 + Uз + ...... + Unr (a + nb) (a + n + 1. b) (a + n + 2 . b). ...... . (a + n + r − 1 . b). Replacing n by n − 1, 1 Un-1 = · (a + n − 1 . b) (a + nb) (a + n + 1. b) ... (a + n + r− 2 . b); .. (a + n + r − 1 . b) u2 = (a + n − 1 . b) Un-1 = vn, say. where C is a quantity independent of n, which may be found by ascribing to n some particular value. Thus SC 1 1 (r − 1 ) b ̊ (a + n + 1 . b) ... (a + n + r−1.b) Hence the sum may be found by the following rule: Write down the nth term, strike off a factor from the beginning, divide by the number of factors so diminished and by the common difference, change the sign and add a constant. V1 The value of C a + rb u; but it is advisable in ch case to determine C by ascribing to n some particular value. A |