Convergence of $\lim_{n\to\infty}(a_n)^n$ What can we say about $\displaystyle\lim_{n\to\infty}(a_n)^n$ given a real sequence $(a_n)$? In particular what happens when $(a_n)$ is convergent? Is it possible for the described limit to converge if $(a_n)$ is not convergent?
 A: $(1,1,1,\ldots)$ converges and $(1^1,1^2,1^3,\ldots)$ converges.
$(2,2,2,\dots)$ converges and $(2^1,2^2,2^3,\dots)$ diverges.
$(\frac12,\frac13,\frac12,\frac13,\ldots)$ diverges and $((\frac12)^1,(\frac13)^2,(\frac12)^3,(\frac13)^4,\ldots)$ converges.
$(2,3,2,3,\ldots)$ diverges and $(2^1,3^2,2^3,3^4,\ldots)$ diverges.
A: We does not say about of (convergence of) $(a_n)$. For example, $\lim_{n\to\infty} n^{1/n} =1$. But $\lim_{n\to\infty} (n^{1/n})^n$ does not converge. But $\lim_{n\to\infty} 1 =1$ and $\lim_{n\to\infty} 1^n=1$.
A: Neither convergence implies (or is implied by) the other.
Consider $a_n=2$ for all $n$. This converges, but the sequence of $a_n^n$ does not.
On the other hand, consider $a_n=\frac12\cdot(-1)^n$. This fails to converge, but $a_n^n$ converges.
A: If a sequence $\{a_n\}_{n\ge1},\{b_n\}_{n\ge1}$ converges then $\{a_nb_n\}_{n\ge1}$  also converges, so for $k\in N$ $\{(a_n)^k\}_{n\ge1}$ also converges if $\{a_n\}_{n\ge1}$ converge but we cant say anything abt the convergence of $\{(a_n)^n\}_{n\ge1}$
That is if the power is also variable(depending on n) then we cant say anything abt. the convergence of the new series.
