Recent content by vivisector

If a_n \le {1 \over {n^2 }} , then \sqrt {a_n } \le {1 \over n} \Rightarrow {{\sqrt {a_n } } \over n} \le {1 \over {n^2 }} ; if a_n \ge {1 \over {n^2 }} , then \sqrt {a_n } \ge {1 \over n} ; multiplying both sides of this inequality by \sqrt {a_n } we obtain...