proof of limit

**tonio** · October 4th 2009, 09:13 AM

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

**redsoxfan325** · October 4th 2009, 10:55 AM

Originally Posted by tonio

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

For large enough $n$ ,

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

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

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

which proves the theorem.

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

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{\displa ystyle\lim_{n\to\infty}a_{n+1}}{\displaystyle\lim_ {n\to\infty}a_n}=\frac{A}{A}=1$

**tonio** · October 4th 2009, 11:23 AM

Huge. thanx a bunch, Redsoxfan

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