# Thread: disprove a convergence question?

1. ## disprove a convergence question?

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

??

2. It is true that $\left( {A_n = \sqrt[n]{n}} \right) \to 1$.
But what about $\left( {\sqrt[n]{n}} \right)^n \to ?$

3. where did you find the power of 1/n
when i do the limit
the base goes to 1 and the power goes to + infinity

that is not solvable

what to do??

they question says prove/disprove

how to disprove

4. Originally Posted by transgalactic
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

$\limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$

Also let us try a three case scenario again

Case #1: $A_n\geqslant A_{n+1}\cdots$

It is clear then from the fact that $A_n\to1$ that there exists some $N$ such that [tex]N\leqslant{n}[/math[ implies

$A_n\geqslant{1}$

From there it is clear then that

$1\leqslant{A_n}\leqslant{A_n^n}$

Or $1\leqslant\sqrt[n]{A_n}\leqslant{A_n}$

Case #2: $A_n\leqslant{A_{n+1}}\cdots$

From here it is clear that there exists a $N$ such that $N\leqslant{n}$ implies $0\leqslant{A_n}\leqslant{1}$

And it should be clear then that $A_n^n\leqslant{A_n}\leqslant{1}$

Or

$A_n\leqslant\sqrt[n]{A_n}\leqslant{1}$

Case #3: This is when $A_n=A_{n+1}\cdots$ where the conclusion readily follows.

5. how did you came to the conclusion that my limit equal this?

$

\limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$

6. Originally Posted by transgalactic
how did you came to the conclusion that my limit equal this?

$

\limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$
Im sorry when I saw Plato's post I mistakenly believed you were asking to prove that $A_n\to1\implies\sqrt[n]{A_n}\to{1}$, forgive me.