344-  5 If lim |a_n|/|a_{n+1}| = R, consider lim |a_{n+1}|/|a_n| =1/R = rho;
        the ratio |a_{n+1}|/|a_n| is within epsilon of rho for n
        greater than K(epsilon) and  |a_K|(rho-epsilon)^{n-K} .lt. |a_n| .lt.
        |a_K|(rho+epsilon)^{n-K}, and we can apply the root test to both sides
        of this inequality.
        The Taylor series for sine has every other term equal to 0, hence this
        limit does not exist. 

      9 The n-th root of any constant approaches 1, hence the radius of 
        convergence is 1.  Alternatively, if x^n blows up (which happens for 
        x .gt. 1) p times x^n blows up, and if x^n goes to zero (which happens
        for |x| .lt. 1) q times x^n goes to zero and the summation is .lt. 
        q times 1/(1-|x|). 

     16 int (1/(1+x)) = log(1+x).  The Taylor series for 1/(1+x) is
        sum ((-1)^n x^n).  integrating termwise and shifting the index of 
        summation provides sum (-1)^n+1 (x^n)/n  log(1) is 0, so the summation
        starts witn n=1.  The radius of convergence is readily verified to be 
        one.