approaches the limit zero as n increases indefinitely (660), and hence for an infinite converging series we have scale of relation 1-p-q, it is only necessary to make x=1 in the results in the preceding article, in order to find the sum of n terms, or of an infinite number of terms. according to ascending powers of x (2670) we shall find that the expansion will be the series u + u1x+u2x2 + whether this series is convergent or divergent. Accordingly, the fraction But the generating function of a recurring series is not the sum of the series unless the series is convergent. 717. General Term of the Recurring Series. If 1 px — qx2 can be resolved into two real factors of the first degree in x, the u+x(u — pu) expression can be decomposed into partial fractions, 1 px qx2 each having for its denominator an expression containing only the first power of x (2688). In this case each partial fraction can be developed into a geometrical progression whose general term is readily found, and the algebraic sum of these general terms will be the general term of the recurring series. EXAMPLE. Find the generating function, the general term, and the sum of the first n terms of the recurring series 7p-q-10, p + 7q43 = 0; whence p = 1, q = 6; and the scale of relation is 1 Hence the generating function is 1 x 6.2 · 1 [1 + (3 x) + (3 x)2 + (3x)3+ . . . + (3 x)”+ . . .]; 2 [1+(−2x)+(−2x)2 +(−2x)3+...+(−2x)”+ Hence the (r+1)th terms of the expansions (3) and (4) are respectively therefore the (r+1)th term of the expansion of the given series (1) is If we add the sums of the first n terms of the series (3) and (4) we shall obtain the sum of the first terms of the given series, namely, 718. The student can readily show that if the series u2 + u1x + u„x2+ is a recurring series of the third order, whose scale of relation is 1 px - qx2 - rx3, and is convergent, the sum of the first n terms is EXERCISE CII NOTE. In the following exercises such values only of x are considered for which the series are convergent. Find the generating function of the following recurring functions: ཇ. 2. 2+5+10+17+ 26x + 37 x + . . . . 3. 1+x+2x2 + 7 x3 + 14 x1 + 35 x2 + . . . . 4. 1+x+2 x2 + 2 x3 + 3 x1 + 3 x3 + 4 x® + 4 x2 + 6. 1*+2*r+3 +4* rẻ tốt t 7. Find the sum of the infinite series 1 + 4 x + 11 x2 + 26 x3 + 57 x1 +120+... when a <. 8. Find the nth term and the sum of the first n terms of the recurring series: (iii). 1 + 5 + 17 + 53 + 161 + 485 + (v). 13x+6.x2+10x+15+. (vi). 4x + 14x2 + 40x3 + 110x1 + 304x3 — 854x® + (vii). 13r+6.x2 + 10.r3 +15.x1 + 21x2 + 28x® + (viii). 2x+2x2 5.x3 + 10.x1 17x+ ... (1 − x)+1 9. Show that the series 1 + 2′′x + 3′′x2 + 4′′x3 + expansion of an expression of the form "+"+.. also that a, = 0 and a-s = @ _ 1° SUMMATION BY UNDETERMINED COEFFICIENTS 719. Thus far it has been shown how to sum three particular kinds of series, namely, arithmetical progression, geometrical progression, and recurring series. The method of undetermined coefficients can be employed to sum a fourth kind of series which usually has the character that the sum of the first n terms of the series is an expression containing a, which expression has the same form no matter what value has. In order to illustrate this method, find the sum of the series where the A's are constants to be determined. Since this equation is to remain true for all values of n, change n into n+1; thus 2 (2) 12+22+33+ . . . +n2+(n+1)2= A ̧ + Â ̧ (n + 1) + A ̧ (n + 1); + A ̧(n + 1)3 + · Subtract (1) from (2), (3) n2 + 2 n + 1 = A ̧ + A ̧ (2 n + 1) + A„(3 n2 + 3n + 1) 2 On equating the coefficients of like powers of n, A, will be 0 since n3 does not occur in the first member of (3); similarly all succeeding coefficients A, A, etc., are zero. Hence, equating coefficients of like powers of n in (3), we have Since this equation is to hold for all positive integral powers of n, put n = 1; then A 0. Hence the required sum is = The same method may be applied to find the sum of the cubes of the first n positive integers or the sum of their fourth powers, and so on. SERIES WHICH MAY BE GIVEN THE FORM OF THE ALTERNATING SERIES (1) 720. It has been shown, 2659, that a series of the type is convergent in case lim (un) = 0 and u+1""" n=x Since the sum of series (1) is u the sum of every series which can be reduced to the form of (1) can be found. 1 1 1 (3) (−x + a) + (x + a ̄ ̄x + 2 a) + ( x + 2 a Hence according to the principle just explained the sum of (2) to infinity is 1 x 721. The sum of the first n terms of a series which can be reduced to the form of (1) can be readily found. For example, the sum of the first terms of (2) is from (3) equal to Show by the method of undetermined coefficients that the sum of: 1. 13+5+ + (2 n 1)2 . n(2n-1) (2n+1) ̧ = 3 n (n+1)(n+2) 3 2. 1.2+2 3+3·44·5 + . . . + n (n + 1) = 4. 1·3 +3·5+5·7 + . . . + (2n −1) (2n+1) = ¦ (4n2 + 6n—1). 6. 1 4 72·5·8+3.6·9 + . . . + n(n + 3) (n + 6) = n(n + 1) (n + 6) (n + 7). 720, 721, sum the following series to n terms and to infinity. 7. 133353 + 73 + . . . + (2n-1)3 n3 (2 no |