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.

Sun Feb 08 1998