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Mathstud28 · #4 December 3rd, 2008, 05:51 PM

Since no one else is trying this I will give a suggestion. This is not a full solution but merely a possible building block for you to work off of. Because of the neccessary montonicity of $a_n$ why not consider three cases: $a_n<a_{n+1}\cdots$ , $a_n>a_{n+1},\cdots$ , or $a_n=a_{n+1}=\cdots$

Case Three needs no explanation.

If case two is true then obviously $\max\left\{a_n,a_{n+1},\cdots\right\}=a_n\to{a}$

And if case one is the case then we would have that $\max\left\{a_n,a_{n+1},\cdots{a_{n^2}}\right\}=a_{n^2}$

So now we just need to use the fact that $a_n\to{a}$ to show that $a_{n^2}\to{a}$

By the definition of $a_n\to{a}$ we have that for every $\varepsilon>0$ there exists a $N$ such that $N\leqslant{n}$ implies $d\left(a_n,a\right)<\varepsilon$

So now because of the above we have that $d\left(a_{n^2},a\right)<\varepsilon$ whenever $N\leqslant{n^2}\implies{\lceil{\sqrt{N}\rceil\leqslant{n}}}$ so $a_{n^2}\to{a}$