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Having trouble figuring out the following problem/proof.

If we let $(a_n)$ be a bounded sequence. I am trying to prove that $(a_n)$ has a subsequence $(a_{n_k})$ with $$\lim_{k\to\infty}a_{n_k}=\limsup a_n$$

I know that the following about a limsup,

Let $(s_n)$ be a sequence in $R$. We define $$\limsup\ s_n = \lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$

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    @Michael Hardy thank you for the edit – 2017-02-10
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    @Michael Hardy would you be able to help me? – 2017-02-10
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    @MichaelHardy yes I typed that in wrong, I have edited my question – 2017-02-10

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Let $s_{N}=\sup\{a_{n}:n>N\}.$ Then we know that for every $\varepsilon>0,$ and for every $N>0,$ there is some $n>N$ such that $s_{N}-\varepsilon\max\{k,n_{k-1}\}$ such that this property holds, i.e., $$s_{k}-1/k

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    are those n k k n letters that are floating supposed to be like $limsup_n$? – 2017-02-10
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    Yes. In inline equations, those appear as subscripts, but when they are displayed, they get tucked underneath, as with summation notation. – 2017-02-10
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    so this proves that I that $(a_n)$ has a subsequence $(a_{n_k}) with $$$\lim_{k\to\infty}a_{n_k}=\limsup a_n$$? – 2017-02-10
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    Yes, the subsequence identified above has this property, since the second set of displayed inequalities force $\lim\inf_{k}a_{n_{k}}=\lim\sup_{k}a_{n_{k}}=\lim\sup_{n}a_{n},$ which means the limit of $a_{n_{k}}$ exists and equals $\lim\sup_{n}a_{n}.$ – 2017-02-10
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    where should I add that into the explanation? could you edit that in – 2017-02-10