disprove a convergence question?

**transgalactic** · December 5th 2008, 02:10 AM

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

??

**Plato** · December 5th 2008, 03:05 AM

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

**transgalactic** · December 5th 2008, 11:38 AM

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

**Mathstud28** · December 5th 2008, 01:11 PM

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

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

And you have already asked this question already.

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.

**transgalactic** · December 5th 2008, 11:38 PM

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

$<br /> <br /> \limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$

**Mathstud28** · December 6th 2008, 08:50 AM

Originally Posted by transgalactic

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

$<br /> <br /> \limsup \sqrt[n]{A_n}=\limsup\frac{A_{n+1}}{A_n}$

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.

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