Now, if n is indefinitely increased, the terms of this series have a finite ratio to the series = Thus the limit, when n∞, of the nth term of the original binomial series is zero: .. the series is convergent. B. Let m+1<0: then the multiplier is arithmetically greater than 1, but decreases down to the limit 1. Now, if n is .. un+1 is infinite. series is infinite. indefinitely increased, this series is infinite; Hence for some values of n = ∞, the binomial Moreover, taking the terms in pairs, we have which is 1 if m + 2 is negative; but, if not, This expression = m + 2 nearly, if r is large enough. But m+ 2 < 1, since m + 1 < 0. Hence, by Art. 451, the series is infinite for all values of n, and is therefore divergent. Thus, when x/a = 1, the series is convergent or divergent according as m is greater or less than - 1. § 2. THE BINOMIAL THEOREM. The index theorem. 463. If, for all positive and negative values of m and n, f(m) and f(n) are single-valued functions of m and n such that ƒ (m) ׃(n) = f (m + n), then, for all such values, f (n) is one of the values of {f(1)}". (1) Let n be a positive integer. × Then f(1) ×f(1) x ... to n factors =ƒ(1+1+... to n terms), i.e. {f(1)}" =f(n). Then ƒ (n) ׃ (0) =ƒ (n + 0) =ƒ (n). ·· ƒ(0) = 1 = {ƒ (1)}o. (3) Let n be a negative integer == m say. Then f(m) xf(− m) = ƒ (m − m) =ƒ (0) = 1 by (2). i. e. ... to q terms 8), (P {ƒ (†)}* = ƒ(p) = {ƒ (1)}P by (1) and (3). is one of the values of Y{ƒ (1)}, i.e. of {ƒ(1)}. 464. When m is a positive integer, it is easily proved by the theory of combinations that m(m−1).....(m−r+1) x2 + ... 1.2 r J. T. 22 Now the series on the right side is endless unless m is a positive integer. It is natural to enquire, then, what is the value of this endless series, when m is other than a positive integer, and when m and x have such values as will secure its convergency, as investigated in Arts. 460–462. Now multiplying the above series by 1 + x, the coefficient of " in the product is m (m − 1)...(m − r + 2) (1 + m− r + 1) = 2 1 r (m+1)m (m-1)... (m-r+2) r Hence for all positive or negative integral values of m, f(m) = (1 + x)m. ...(I). .(II). By means of the index theorem we will find the value of f(m) for any positive or negative value whatever. 465. Lemma. Lemma. If two sets of quantities So, S1, ...S, and To, T1, ...Tr+1 are so related that, for any suffix P, then (m − r + p) Sp = (r − p + 1) T2 and (n−p+1) Sp-1=pTp; (m + n − r) (sum of the S's) will equal (r + 1) (sum of the T's). In the first of the given equations, put p=0: Thus (mr) S(+1) To. In the last of the given equations, put pr+1: Thus and so on. (m + n − r) S2 = (m − r) S2+nS。 = (+1) To+... − 1) Sp-, + (n-p+1) Sp-1...+pTp, p) Sp + (n − p) Sp = (r− p + 1) Tp (m + n − r) S, = mS, + (n − r) S, ...+(+1) Tr+1 == + Adding all these equations together, we see that the coefficient of Tp is r + 1 for every value of p. Hence (m + n − r) (sum of the S's) = (r+ 1) (sum of the T's) 466. To prove universally that the series Q.E.D. for all positive or negative values of m for which it is convergent, Hence, multiplying the two series ƒ (m), ƒ (n), coeff. of x in f(m) × ƒ (n) = S。 + S1 + S2 + ... + Sr, 2 coeff. of +1 in ƒ (m) × ƒ (n) = T + T1 + T2+ ... + Tr+ Tr+19 .. by lemma, coeff. of Now coeff. of xo = 1. coeff. of x2+1 in ƒ(m) × f(n) = Hence, for all Now f(1) = 1 + x. 0 1 +1 (coeff. of x") × m + n − r r+1 Hence putting r=0, 1, 2... successively. (m + n) (m + n − 1)... (m + n − r) r+1 f(m) ׃(n) =ƒ (m+n). real values of m, f(m) is one value of {ƒ(1)}". .. the series f(m) is one value of (1 + x)m. 467. The above series, which involves m and x, may be called f(m, x). If now m is integral, (1+x)m has but one value. Hence we may say f(m, x) = (1 + x)m. = But if m is fractional say p/q, (1+x)m has a number (at present unknown) of values. It has been proved, in Art. 230, however, that the qth root of a positive quantity has only one positive value. Now for all values of f(m, x) which are convergent, x is numerically not greater than 1, so that for all such values 1 + x is positive. We will proceed to show that for all positive or negative values of m and x, which make ƒ(m, x) endless but convergent, f(m, x) is positive. (I) Let x and m be both negative. Then every term in ƒ (m, x) is positive: .. ƒ (m, x) is positive. (II) Let x be negative, and m positive. Then f(m, x) = f(n, x) × ƒ (m −n, x). Suppose n is any integer greater than m. |