Suppose $t$ and $n$ are positive integers. What is the best lower bound we can have for the following term? $$1-(1-\frac{1}{n})^{2t} \ge ?$$ So far I am using $f(x):=x^{2t}$ is convex, and thus $$1-(1-\frac{1}{n})^{2t}=f(1) - f(x) \ge \nabla f(x) (1-x)=\frac{2t}{n}(1-\frac{1}{n})^{2t-1}.$$ But the RHS goes to 0 when $t$ goes t0 0 while the LHS converges to 1.
Many thanks.