$A=\{\frac{n+(-1)^n}{2n}:n\in\Bbb N,n>1\}$; find and prove $\inf A$

Question

Let $$A=\left\{\frac{n+(-1)^n}{2n}:n\in\Bbb N,n>1\right\}$$ Find $\inf A$ and prove that it's the infimum.

I believe $\min A=\frac13$. Of course I know that if the $\min$ exists, $\inf$ exists. How can I prove this?

Answer 1

Let $a_n = \frac{n+(-1)^n}{2n}$, then $$a_{2n}=\frac{2n+1}{4n} = \frac12 + \frac1{4n}\geqslant \frac12$$ for all $n$. Considering the odd terms, we have $$a_{2n+1}=\frac{(2n+1)-1}{2(2n+1)} = \frac12 - \frac1{2(2n+1)}\geqslant \frac12 -\frac1{2(2\cdot 1+1)}=\frac13=a_3.$$ Since $$A=\left\{a_{2n}:n=1,2,\ldots\right\}\cup\left\{a_{2n+1}:n=1,2,\ldots\right\}, $$ it follows that $\min A = \frac13$.