I've got a number sequence, $ a_n = \frac{1+(-1)^n 2n}{1+3n} $ and I have to calculate the upper limit and the lower limit.
First of all, I've divided the sequence in two subsequences: the even subsequence {$a_{2n}$} and the odd subsequence {$ a_{2n+1} $}.
Furthermore, it's obvious that $ a_{2n}>a_{2n+1} $ , owing to the fact that all the terms of $ a_{2n} $ are positive and all the terms of $a_{2n+1}$ are negative.
So I would like to know if it's enough to say that $\lim_{x\to \infty}inf(a_n)=\lim_{x\to \infty}(inf_{k\ge n}(a_k))=\lim_{x\to \infty}(inf(a_k,a_{k+1},a_{k+2},...))=\lim_{x\to \infty}a_{2n+1}=-\frac{2}{3}$
No, it is not correct, but it is almost correct. For each $k\in\Bbb N$, you have$$\inf\{a_k,a_{k+1},a_{k+2},\ldots\}=-\frac23,$$since each $a_k$ is greater than $-\frac23$ and $\lim_{k\to\infty}a_{2k-1}=-\frac23$. So, you can put it as follows:\begin{align}\liminf_{k\to\infty}a_k&=\lim_{k\to\infty}\inf\{a_k,a_{k+1},a_{k+2},\ldots\}\\&=\lim_{k\to\infty}-\frac23\\&=-\frac23.\end{align}