Find $\lim_{k \rightarrow \infty} \sup a_k$ and $\lim_{k \rightarrow \infty} \inf (a_k)$, with $a_k=(\frac{1}{k})_{k \in \mathbb{N}}$

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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?

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Yes, this answer is correct. The limit superior and limit inferior of $\{a_k\}$ are equal $\iff$ $\lim_{k\rightarrow\infty}a_k$ exists, in which case all three limits are equal.

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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 $\lim\sup\limits_{n\to\infty} x_n=\lim\inf \limits_{n\to\infty}x_n=0$