= u-run-i =0; the sum of the coefficients of u and Un- with their proper signs, that is, 1-r, is called the scale of relation. 3хи ...... Again, in the series 2+4x+14x2 + 46x3 + 152x* + the law connecting consecutive terms is u-3xu-1-x3u-, = 0; this law holds for values of n greater than 1, so that every term after the second can be obtained from the two terms immediately preceding. The scale of relation is 1-3x-x. 651. To find the sum of n terms of a recurring series. Let the series be u ̧+u ̧x+u ̧x2 +Û2x3 + ................, and let the scale of relation be 1-px- qx3, so that for every value of n greater than unity u - pu2- 1 — qu„ _ ̧ = 0. Denote the first n terms of the series by S, then 2-1 for all the other terms on the right-hand side disappear by virtue of the relation which holds between any three consecutive terms of the given series; therefore diminishes without limit, we may say that the sum of an infinite number of terms of the recurring series is u+x (u, -pu) It is obvious, that if this expression be developed in a series according to powers of x, we shall recover the given recurring series. (See Art. 648.) ....... and 652. If the recurring series be u+u ̧+u ̧+U; + the scale of relation 1-p-q, we have only to make x = 1 in the results of the preceding Article, in order to find the sum of n terms, or of an infinite number of terms. 653. The expression u12+x (u1-pu) may sometimes be de1 - px - qx3 composed into partial fractions, each having for its denominator an expression containing only the first power of x (see Art. 643). When this can be done, since each partial fraction can be developed into a geometrical progression, we can obtain an expression for the general term of the recurring series. We have thus also another method of obtaining the sum of n terms, since the sum of n terms of each of the geometrical progressions is known. EXAMPLES OF RECURRING SERIES. Find the expressions from which the following three series are derivable; resolve the expressions into partial fractions, and give the general term of each series. 4. Find how small x must be in order that the series in Example 3 may be convergent. 5. Find the general term of the series 3+11 +32 + 84 + ..... 6. Sum the following series to n terms, 1+5+17 +53 + 161 + 485 + ...... 7. Find the general term of the series 10+14 + 10 + 6 + ... and the sum to infinity. 8. Find the expression from which the following series is derivable, and obtain the general term 2-a+2a2-5a3 + 10a1 — 17a3 + L. SUMMATION OF SERIES. 654. Series of particular kinds have been summed in the Chapters on arithmetical progression, geometrical progression, and recurring series; we shall here give some miscellaneous examples which do not fall under the preceding Chapters. 655. To find the sum of the series 1+ 2+ 33 +...... + n3. We have already found this sum in Art. 482; the following method is however usually given. Assume 1a + 2a +32 + +n2 = A + Bn + Cn3 + Dn3 + En1 +......, ...... By subtraction, + n2 + (n + 1)3 = A + B (n + 1) + C (n + 1)2 + D (n + 1)3 + E (n + 1)*+. n2 + 2n + 1 = B + C (2n + 1) + D (3n2 + 3n+1) +E (4n3 + 6n3 + 4n + 1) + Equate the coefficients of the respective powers of n; thus E = 0, and so any other term after E would = 0; 3D=1; 3D+20=2; D+C+B=1; To determine A we observe that since this equation is to hold for all positive integral values of n, we may put n=1; thus A=0. Hence the required sum is The same method may be applied to find the sum of the cubes of the first n natural numbers, or the sum of their fourth powers, and so on. See also Art. 666. 656. Suppose the nth term of a series to be {an + b} {a(n + 1) + b} {a (n + 2) +b} {a (n + m − 1) + b}, where m is a fixed positive integer, and a and b known constants; then the sum of the first n terms of this series will be {an + b} {a (n + 1) +b} ...... {a (n + m − 1) + b} {a (n + m) + b} +C, (m + 1) a where C is some constant. Let u, denote the nth term of the proposed series, S the sum of n terms; then we have to prove that an + b (m + 1)a n+1 и + C. Assume that the formula is true for an assigned value of n; add the (n + 1)th term of the series to both sides; then thus the same formula will hold for the sum of which was assumed to hold for the sum of n terms. n + 1 terms, Hence if the formula be true for any number of terms it is true for the next But the formula will be true when greater number; and so on. n = 1 if we take C such that thus C is determined and the truth of the theorem established. Thus the sum of the first n terms of the proposed series is ob tained by subtracting the constant quantity certain expression which depends on n. an + b (m + 1) a (m + 1) a This expression is uwe may also put this expression into the equi valent form observe that it can be formed by introducing an additional factor at the end of u, and dividing by the product of the number of factors thus increased and the coefficient of n. 657. We may obtain the result of the preceding Article in another way. As before, let u, denote {an + b} {a (n + 1) + b} {a (n + 2) +b} {a (n + m − 1) + b}, ...... and let S denote the sum of the first n terms of the series of 12 |