Question 933035
True for n = 1


Assume *[tex \large \frac{1}{n} > \frac{1}{2^n}] for some fixed *[tex \large n \ge 1]


Then *[tex \large \frac{1}{2n} > \frac{1}{2^{n+1}}] by multiplying both sides by 1/2. However *[tex \large \frac{1}{n+1} \ge \frac{1}{2n}] since n is at least 1. It follows from transitivity that *[tex \large \frac{1}{n+1} > \frac{1}{2^{n+1}}] so the statement is true for all *[tex \large n \ge 1].