2.7311 r 211 7. Find the sum of x terms of the series of products 2011 7x50 10x415 19.2 +2 3+39.4 +45 + &c. Ans. + 3 8. Find the sum of x terms of the series of products 2411 7311 ..3 +2.5+3.72 +4.99 +5.11°+&c. Ans. + + 4 3 2 9. Find the sum of x terms of the series of products 1.2.3+2.4.6+3.6.9+4.8.12+ &c. as also the sum of ten terms of 3.7411 the same series. Ans. -2311 + 8x21. 2 x411 3.x211 2 11. Find the sum of ten terms of the series -3311 + + +&c. 3'+3+3+3 +35+ &c. Ans. 4 2 12. Find the sum of ten terms of the series of fractions 1 1 1 + 2.3.4 3.4.5.4.5.6 1 Here t = = (x+1, -31, -1 -2 = 2(x+2) 2/1 1 : 2.2211 2.2211 (247-12) 2.12011 Make I=0, then will whence it appears 2(x + 2)211 1 that the integral is too small by 2.2217", which, therefore, must be 1 added to it; and we shall have E Now 2(2+2)211 1 1 1 make x = 10, and we shall have 2.2211 for the sum of ten terms. 1 Scholium. Here it is evident that is the sum of the infinito series ; we may, therefore, find the sum of an infinite series of fractions, by the following Rule. Transfer the denominator to the numerator, and the exponent will become negative ; add unity to the exponent, and divide the fraction by the exponent thus increased, and by the difference, and the fraction thus divided, will be the sum of the infinite series. Observing in dividing by the negative exponent-to change its sigo into t. 2.2017 Examples 1 1 1 .. Find the sum of the Infinite series + + +&c. + the sum. 1 3. Find the sum of the infinite series 1 + &c. =3-413. 1 9 9.3.6.9 Scholium. When the numerators contain any powers or products of the distance, with or without constant quantities, the integral will be easily found by the following Rule. Take the differences of the numerators of the given fractions until the last order became constant : then the first numerator of the given fractions, the first of the first, the first of the second, the first of the third, &c. order of differences are the respective numerators of the integrals ; find as many orders of the integrals of the respective deriominators as the numerators so found, and these integrals will be the respective denominators of the fractions, which will be the value of the sum. Examples i Find the sum of the infinite series atb a+26 @+31 + (m + &c. Here the numerators are a+b, a+21, a +36, &c. and their first differences are b, b, Now a'm-se m-21 Numerators 4, 9, 16, First differences, 5, 7, Second differences, 2, constant. Then as many of the succeeding values of the denominators T 1 1 1 1 (-1) 2 Whence = + + ON THB CONVERGENCY OF SERIES. Problem 1. To find the convergency of an Infinite Series in general terms, x being the distance of the term from the beginning. Find the factors in terms of 7, the factors or variable quantities being supposed to be in arithmetical progression; then multiply into, or divide by the constant quantities, will give the general term, which reserve; then form the preceding term in the same manner, by writing r-1 for I; divide the subsequent term by the preceding; and the quotient will shew the convergency of the series for any value of x. Examples. 1. To find the convergency of the infinite series 1 1 mots na • 1 + n! nina 1 therefore which is the convergency roquired, and is in nua this example a constant quantity. 2. To find the convergency of the infinite series 1 1 1 1 1 + + 1 2 3 4 .... The preceding term is –,; therefore the convergency is * ! 3. To find the convergency of the infinite series 4. To find the convergency of the infinite series 1 1 1 1 it + amid (a tálald (a +2d) mid t. 1 The preceding term is ; therefore the convergency is (242 (2-2)ml? ( ->)zm-113 2–18 [z+?(m-1)] z+z(m-1) 1 1 1 Here the general term is r»[4(2-1)+1]; or, putting a for the first factor, and c for the increment, it will be and the gru 1)' 1 *1th term will be 5.38't for the convergency c(-2+a term by the I-1th term, gives ra[c(2-1)+a] of the series. Scholium. Letx=2; then the convergency c(1-2)+a pre [c(x - 1) + a) becomes simply mic4a) · Whence, if the first factor a be increased, a (ca the convergency is less ; that is to say, supposing med and c to be con stant, the greater a is, the nearer will the expression pe (c+a) approach to which is its limit ; and if r=1, this limit wouid be unity. 6. Find the convergency of the infinite series m21102 + &c. R?pm+116 Ropm+2! Ropm+ 3le .) This series is equivalent to тся (m+1)CB Romic R(P+mc) * (+(m+1)x)+ &c. Therefore the con mc + 1 |