Find $\lim_{k \rightarrow \infty} \sup a_k$ and $\lim_{k \rightarrow \infty} \inf (a_k)$, with $a_k=(\frac{1}{k})_{k \in \mathbb{N}}$ Find $\lim_{k \rightarrow \infty} \sup a_k$ and $\lim_{k \rightarrow \infty} \inf (a_k)$, with $a_k=(\frac{1}{k})$
Per definition: $\lim_{k \rightarrow \infty} \sup(a_k) = \lim_{k \rightarrow \infty} \sup (\bigcup_{i=k}^\infty \{a_i\})$
I have $\sup (\bigcup_{i=k}^\infty \{a_i\}) = \frac{1}{k}$
$\Rightarrow \lim_{k \rightarrow \infty} \sup (\bigcup_{i=k}^\infty \{a_i\}) = 0$
For the limes inferior I get:
$\lim_{k \rightarrow \infty} \inf(a_k) = \lim_{k \rightarrow \infty} \inf (\bigcup_{i=k}^\infty \{a_i\})$
Since $\inf (\bigcup_{i=k}^\infty \{a_i\}) = 0$
$\Rightarrow \lim_{k \rightarrow \infty} \inf(a_k) = 0$
Is that correct? Can it be that the limes superior and limes inferior of a sequence are equal?

We know that a bounded sequence$(x_n)_{n=1}^{\infty}$ is convergence if and only if $\lim\sup\limits_{n\to\infty} x_n=\lim\inf \limits_{n\to\infty}x_n$. Since $(a_{k})=(\frac{1}{k})_k$ converges to $0$, so that $\lim\sup\limits_{n\to\infty} x_n=\lim\inf \limits_{n\to\infty}x_n=0$
