Given f(x) = xe-x2 I can differentiate once and use Leibniz to show that for n greater than 1 f(n) = -2nf(n-2) - 2xf(n-1) I want to show that the Maclaurin series for f(x) converges for all x. At x = 0, the above Leibniz formula becomes f(n) = -2nf(n-2) I know that f(0) = zero so this implies that even terms of Maclaurin series are zero, whilst f(1)(0) = 1, f(3)(0) = -6*1 = -6, f(5)(0) = -10*-6*1 = 60 and so on. I assume that to show convergence, I need to find a formula for the nth term of the Maclaurin series, then use the ratio test to show that terms are decreasing. I can see the pattern (the derivative increases by a factor of 2(2n + 1) each time) but am unsure how to express this in a formula. Perhaps using factorials? Thanks for any help.