307 -5 1 = 1 - (-x) + (-x) - x^2 + x^2 + ... - (-x)^n + (-x)^n = (1 - x + x^2 - x^3 + ... + (-x)^(n-1))(1+x) + (-x)^n and divide both sides by (1+x) Take the integral of both sides from 0 to x. Move a bunch of stuff to the left of the equal sign; the inequality follows from the fact (1/(1+t) .lt. 1/1) and the triangle inequality. 15 log_a (x) = log(x)/log(a) ; log(a) is defined for a .gt. 0 and nonzero for a .neq. 1, log(a) is a constant wrt x. Hence log_a (x) is differentiable wherever log (x) is. the derivative is just (1/log(a) D(log(x)) ) = (1/log(a))(1/x). 322 -6 1/((a + n)(a + n +1)) = 1/(a+n) - 1/(a+n+1). This provides a telescoping series. For finite summation up to k, it is equal to 1/a - 1/(a+k+1) as k goes to infinity, 1/(a+k+1) goes to 0, which provides the result. 1/(n(n+1)(n+2)) = 1/2n + -1/(n+1) + 1/2(n+2). This is telescoping and the tail terms go to zero, but because the cancellation is of -1/(n+1) with the previous 1/2(n+2) and the next 1/2n, the lead terms 1/(2 times 1) - 1/(1 + 1) + 1/(2 times 2) remain, and this sum equals .25 9 By regrouping the double summation, we se we have a_1 (1+ 1/2 + 1/3 + 1/4 + ...) + a_2 (1/2 + 1/3 + 1/4) + etc. and since the a_i are all positive and the expressions in parentheses are infinite, the summation is infinite. 18 SUM (1/(1 + a^n)) is less than SUM (1/a^n) which converges for a .gt. 1 hence the sum is finite for a .gt. 1. If a .lt. 1, 1/(1 + a^n) approaches 1 for large n, i.e., the tail terms are near 1, hence the series diverges.

Sun Jan 25 1998