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I am wondering if this is true:

If $\{a_n^k\}_{n=1}^\infty\rightarrow a^k$, and $\{a^k\}_{k=1}^\infty\rightarrow a$, does then $\{a_n^n\}_{n=1}^\infty\rightarrow a$?

I tried proving it but I got stuck. Given an $\epsilon$ I can find a K for $\{a^k\}_{k=1}^\infty$ such that $k \ge K \rightarrow |a^k-a|<\epsilon/4$. But that is as far as I get, since for the first sequence I can only find an N(k), but I need this N to hold for infinitely many K's.

Is the statement false?, is there a counter-example?

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It's false. Consider $$a_n^k=\begin{cases} 1 & \text{if }n\leq k,\\ 0 & \text{if } n>k \end{cases}$$ so that $\{a_n^k\}=\{1,1,\ldots,1,0,0,\ldots\}$ with $k$ $1$'s. Then $\{a_n^k\}\to 0$ for all $k$, so that $a^k=0$, and hence $a=0$, but the sequence $\{a_n^n\}=\{1,1,\ldots\}$ does not converge to $0$.

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Simplest counterexample, in my opinion:

$$a_n^k = \cases{1&if $n=k$\\0&otherwise}$$

$\lim\limits_{n\to\infty}a_n^k=a^k=0$ for all $k$, so $a=0$. But $a_n^n\to1$, clearly.

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