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.

 

 

Russell Campbell
Sun Jan 25 1998