21. Shew that the sum of the squares of all the numbers less than a given number N and prime to it is a, b, c... being the different prime factors of N. 22. If p and q are any two positive integers, shew that pq is divisible by (P). [q and by (q)". P. 23. Shew that the square numbers which are also triangular are given by the squares of the coefficients of the powers of x in the expansion of and that the square numbers which are also pentagonal by the coefficients of the powers of x in the expansion of 1 1-6x+x2' 1 1-10x+x2* 24. Shew that the sum of the fourth powers of all the numbers less than N and prime to it is a, b, c,... being the different prime factors of N. 25. If (N) is the number of integers which are less than N and prime to it, and if x is prime to N, shew that x(N) – 1 = 0 (mod. M). 26. If d1, da, dz,.. denote the divisors of a number N, then Shew also that $(d1)+(d2)+(d)+...N. ... *CHAPTER XXXI. THE GENERAL THEORY OF CONTINUED FRACTIONS. *436. In Chap. xxv. we have investigated the properties of Continued Fractions of the form where a, a,.** 1 1 + are positive integers, and a, is either a positive integer or zero. We shall now consider continued fractions of a more general continued fraction. We shall confine our attention to two cases ; (i) that in which the sign before each component is positive; (ii) that in which the sign is negative. *438. To investigate the law of formation of the successive convergents to the continued fraction We see that the numerator of the third convergent may be formed by multiplying the numerator of the second convergent by а39 and the numerator of the first by b, and adding the results together; also that the denominator may be formed in like manner, 3 Suppose that the successive convergents are formed in a similar way; let the numerators be denoted by P1, P2, P3......., and the denominators by 9,, X2, X3, ·· ... Assume that the law of formation holds for the nth convergent; that is, suppose PaPbP-2 91 = a+b2In-2 R The (n+1)th convergent differs from the nth only in having a + b +1 in the place of a„; hence b a Pa+1 = a+1 P+n+1Pn−1 + 1 = a+ 12 + b2+ 19 n − 1) we see that the numerator and denominator of the (n + 1)th convergent follow the law which was supposed to hold in case of the nth But the law does hold in the case of the third convergent ; hence it holds for the fourth; and so on; therefore it holds universally. *439. In the case of the continued fraction a result which may be deduced from that of the preceding article by changing the sign of b„. and is therefore a proper fraction; hence P+P is numerically less than PP-1, and is of opposite sign. In In-1 9+1 In By reasoning as in Art. 335, we may shew that every convergent of an odd order is greater than the continued fraction, and every convergent of an even order is less than the continued fraction; hence every convergent of an odd order is greater than every convergent of an even order. Also Pen-1-Pen is positive and less than P-1P-2; hence 12n-1 I21-2 Hence the convergents of an odd order are all greater than the continued fraction but continually decrease, and the convergents of an even order are all less than the continued fraction but continually increase. Suppose now that the number of components is infinite, then the convergents of an odd order must tend to some finite limit, and the convergents of an even order must also tend to some finite limit; if these limits are equal the continued fraction tends to one definite limit; if they are not equal, the odd convergents tend to one limit, and the even convergents tend to a different limit, and the continued fraction may be said to be oscillating; in this case the continued fraction is the symbolical representation of two quantities, one of which is the limit of the odd, and the other that of the even convergents. The continued fraction will have a definite value when n is a+1 is greater than zero so also is the limit of also neither of these terms can be negative; hence if the limit of n+1 case the limit of Pn+1 Pn bn+12n-1 is the limit of the product of an infinite number of proper fractions, and must therefore be equal to zero; that is, and Pa tend to the same limit; which proves the proposition. For example, in the continued fraction n2 In 12 22 32 3+ 5+ 7+ 2n + 1 + and therefore the continued fraction tends to a definite limit, |