infimum (and supremum) proof

Question

I'm trying to find and prove the infimum of the set:

$X = ((-1)^n)* 1/(2n+1)$

I assumed that the infimum is :

$-1/3$

and proved it like this:

1.$((-1^n)* 1/(2n+1))> -1/3$

(when n is odd),after some math I got that

$n>=1$

2. I needed to show that -1/3 is the biggest, so:

$(-1* 1/(2n+1)) < -1/3 + \epsilon$

and I got that:

$n>= (3*\epsilon -2)/(6*\epsilon -2)$

I am not sure about the second part of my proof, Can you check if it is good ,and explain how to prove the second part of infimum and supremum (to show that the number is the biggest or the lowest)

Thanks

Answer 1

$a_n:= (-1)^n\dfrac{1}{2n+1},$ $n=0,1,2,3,.....$.

$a_1= -1/3.$

Left to show: $a_1 \le a_n$ , $n= 1,2,3,4,...$.

1) For even $n$: $a_n \gt 0 \gt a_1 = -1/3.$

2)Let $n$ be odd: $n=2k+1$, $k= 1,2,3,....$

$\dfrac{1}{2(2k+1) +1} =$

$ \dfrac{1}{4k+3} \lt \dfrac{1}{3}$, or

$-\dfrac{1}{3} \lt -\dfrac{1}{4k+3} =a_{2k+1}$.

Hence :

$-\dfrac{1}{3} \le a_n , n=1,2,3.....$.

Finally :

$\min${$a_n| n =1,2,3..$} $=-1/3.$