Let {h_n} be a sequence of function defined on the interval (0,1) where h_n(x) = (n+n)x^(n-1)(1-x) a. find lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx. b. show that lim (n-> +oo) h_n(x) = 0 on (0, 1) c. Show that lim (n-> +oo) (integral) (from 0 to 1) h_n(x) dx is not equal to integral (from 0 to 1) (0 dx). What went wrong? SOlutions: a. lim (n-> +oo) integral (from 0 to 1) h_n(x) dx = lim (n-> +oo) integral (from 0 to 1) (n+n)x^(n-1)(1-x) dx =lim (n-> +oo)(n+n) integral (from 0 to 1)x^(n-1)(1-x) dx = lim (n-> +oo)(n+n) (1/n - 1/(n+1)) = lim (n-> +oo)n(n + 1) (1/((n)(n+1)) = 1. b. I used the n-th term test in proving this... because if the series of h_n(x) is convergent then lim (n-> +oo) h_n(x) = 0 on (0, 1). But by ratio test, h_n(x) is convergent because the limit of a(n+1)/a(n) as n -> +oo is x, but 0 < x < 1. c. That's the part that I got stuck... well, it seems that the statement above is true... how do I solve this?
Roughly speaking, the main contribution to the integral for large n comes from an increasingly small neighborhood near 1. For any given point x, no matter how close to one, h_{n}(x) eventually gets very small as n increases. But there are still points left between x and 1 for which the value of h_{n} evaluated at these points is very large, and this keeps the value of the integral at one. By the way, did you mean n+n^2 for the factor in front of the function?