Can you help me with any of these? For any natural number n > 1, prove that (4n) / (n + 1) < [(2n)!] / [(n!)2]. For any natural number n > 1, prove that 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(n) > sqrt(n). For any natural number n and any x > 0, prove that xn + xn - 2 + xn - 4 + ... + x-n >= n + 1. Thanks, John 10th Grader Hi John, I'll do problem 2. You first ned to verify that the statement is true when n = 2. 1/sqrt(1) + 1/sqrt(2) = 1 + 1/sqrt(2) > 1.7 > sqrt(2) Now assume that the statement is true for some integer k which is greater than 2. That is 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) > sqrt(k).   * You are assuming that this statement is true for the specific integer k. What you need to do is to show that 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) + 1/sqrt(k+1) > sqrt(k+1) is also true. You can use the asumption * above to see that 1/sqrt(1) + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k) + 1/sqrt(k+1) > sqrt(k) + 1/sqrt(k+1) = (sqrt(k)sqrt(k+1) + 1)/sqrt(k+1) = (sqrt(k(k+1)) + 1)/sqrt(k+1) = (sqrt(k2 + k) + 1)/sqrt(k+1) > (sqrt(k2) + 1)/sqrt(k+1) = (k+1)/sqrt(k+1) = sqrt(k+1) This completes the induction step and hence 1/sqrt(3) + ... + 1/sqrt(n) > sqrt(n) is true for all n > 1. Cheers, Penny Go to Math Central