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Find limit superior and limit inferior of $\frac{n+(-1)^n n^2}{n^2+1}$: the subsequential limit is $1$ and $-1$. so the limit $\sup = 1$ and limit $\inf = -1$ let $E_k = \{a_n\mid n>k\}$.

$$E_k= \begin{cases} \dfrac{k-k^2}{k^2+1},\dfrac{(k+1)+(k+1)^2}{(k+1)^2+1},\dots, \quad \text{if $k$ is odd} \\[6pt] \dfrac{k+k^2}{k^2+1},\dots, \quad \text{if $k$ is even} \end{cases} $$

so $\inf E_k = -1$ and

$$\sup E_k = \begin{cases} \dfrac{(k+1)+(k+1)^2}{(k+1)^2+1} \text{if $k$ is odd}\\[6pt] \dfrac{k+k^2}{k^2+1} \text{if $k$ is even} \end{cases}$$

therefore $\liminf a_n = -1$ and $\limsup a_n = 1$

is this proof right? if not any hint.

Martin Sleziak
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jason
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  • please approve massive edit and clarify that last line. Also three days is enough time to start using Latex on your questions. –  Oct 22 '14 at 02:03
  • Apparently you wrote $E_k={a_n\mid n>k}$ and someone changed it to $E_k=a_n\mid n>k$. The former is a set depending on $k$, not a number depending on $k$. The line below it is not as clear that $E_k$ is a set rather than a number. ${}\qquad{}$ – Michael Hardy Oct 22 '14 at 02:24
  • Looks good now. – Quang Hoang Oct 22 '14 at 02:25

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