Hi:

how could one prove that if a_n --> A and A != 0 then

a_(n+1)/a_n --> 1 ?

I think some average inequalities could help here but I can't see it.

Thanx

Printable View

- Oct 4th 2009, 09:13 AMtonioproof of limit
Hi:

how could one prove that if a_n --> A and A != 0 then

a_(n+1)/a_n --> 1 ?

I think some average inequalities could help here but I can't see it.

Thanx - Oct 4th 2009, 10:55 AMredsoxfan325
For large enough $\displaystyle n$,

$\displaystyle |a_{n+1}-A|<\epsilon\implies\left|\frac{a_{n+1}}{a_n}-\frac{A}{a_n}\right|<\frac{\epsilon}{a_n}$

Taking $\displaystyle n\to\infty$ (and noting that $\displaystyle a_n\to A$) gives us:

$\displaystyle \left|\lim_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\r ight)-\frac{A}{A}\right|<\frac{\epsilon}{A}\implies\left |\lim_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\right)-1\right|<\frac{\epsilon}{A}$

which proves the theorem.

An easier, though less technical way of doing is noting that

$\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{\displa ystyle\lim_{n\to\infty}a_{n+1}}{\displaystyle\lim_ {n\to\infty}a_n}=\frac{A}{A}=1$ - Oct 4th 2009, 11:23 AMtonio
Huge. thanx a bunch, Redsoxfan