i know that An->1
i need to prove that (An)^n ->1
but when i construct limit
lim (An)^n
n->+infinity
i get 1^(+infinity) which says that there is no limit
what do i do in this case in order to disprove that (An)^n->1
??
Why not try this method? Your book ought to show that the Root and Ratio test always yield the same result. So
$\displaystyle \limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$
And you have already asked this question already.
Also let us try a three case scenario again
Case #1: $\displaystyle A_n\geqslant A_{n+1}\cdots$
It is clear then from the fact that $\displaystyle A_n\to1$ that there exists some $\displaystyle N$ such that [tex]N\leqslant{n}[/math[ implies
$\displaystyle A_n\geqslant{1}$
From there it is clear then that
$\displaystyle 1\leqslant{A_n}\leqslant{A_n^n}$
Or $\displaystyle 1\leqslant\sqrt[n]{A_n}\leqslant{A_n}$
Case #2: $\displaystyle A_n\leqslant{A_{n+1}}\cdots$
From here it is clear that there exists a $\displaystyle N$ such that $\displaystyle N\leqslant{n}$ implies $\displaystyle 0\leqslant{A_n}\leqslant{1}$
And it should be clear then that $\displaystyle A_n^n\leqslant{A_n}\leqslant{1}$
Or
$\displaystyle A_n\leqslant\sqrt[n]{A_n}\leqslant{1}$
Case #3: This is when $\displaystyle A_n=A_{n+1}\cdots$ where the conclusion readily follows.