# Finding indexes.

by The_Iceflash
Tags: indexes
 P: 50 I for the most part have this completed but I have a small question and thus checking if I did this correctly. 1. The problem statement, all variables and given/known data Given the Sequence = $$\frac{n}{n+2}$$ $$\approx_{\epsilon}$$ 1 , for n >> 1 Show what index if $$\epsilon$$ = .001 " " if $$\epsilon$$ = .000002 " " for any $$\epsilon$$ > 0 2. Relevant equations N/A 3. The attempt at a solution for $$\epsilon$$ = .001 I did: $$\left|\frac{n}{n+2}-1\right|< .001$$ $$\left|\frac{-2}{n+2}\right| < .001$$ $$\frac{2}{n+2} < .001$$ $$\frac{2}{n+2} < \frac{1}{1000}$$ $$\frac{n+2}{2} > 1000$$ $$n+2 > 2000$$ $$n > 1998$$ The only issue I have with this one and the next one is that they aren't decreasing and I'm not sure if it needs to be or not: $$\frac{1999}{1999+2} = .9990004998$$ $$\frac{2000}{2000+2} = .9990009990$$ for $$\epsilon$$ = .000002 I did: $$\left|\frac{n}{n+2}-1\right|< .000002$$ $$\left|\frac{-2}{n+2}\right| < .000002$$ $$\frac{2}{n+2} < .000002$$ $$\frac{2}{n+2} < \frac{2}{1000000}$$ $$\frac{n+2}{2} > \frac{1000000}{2}$$ $$n+2 > 1000000$$ $$n > 999998[tex] for any [tex]\epsilon$$ > 0 $$\left|\frac{n}{n+2}-1\right|< \epsilon$$ $$\left|\frac{-2}{n+2}\right| < \epsilon$$ $$\frac{2}{n+2} < \epsilon$$ $$\frac{n+2}{2} > \frac{1}{\epsilon}$$ $$n+2 > \frac{2}{\epsilon}$$ $$n > \frac{2}{\epsilon}-2$$
Mentor
P: 21,216
 Quote by The_Iceflash The only issue I have with this one and the next one is that they aren't decreasing and I'm not sure if it needs to be or not:
Who are "they" in "they aren't decreasing"? The terms in the sequence n/(n + 2) are increasing, but you're looking at the difference n/(n + 2) - 1. This sequence is decreasing.
P: 50
 Quote by Mark44 Who are "they" in "they aren't decreasing"? The terms in the sequence n/(n + 2) are increasing, but you're looking at the difference n/(n + 2) - 1. This sequence is decreasing.
The index:

I got n > 1998 as the index so,

$$\frac{1999}{1999+2} = .9990004998$$

$$\frac{2000}{2000+2} = .9990009990$$

 Mentor P: 21,216 Finding indexes. This is to be expected in a sequence that is increasing, as {n/(n + 2)} is. A larger index gives you a larger value.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,292 As Mark44 said, that is to be expected. $\epsilon$ being smaller means you must get closer to the limit value which, in turn, means you must go further out in the sequence. As $\epsilon$ gets smaller, you should expect the index, N, to get larger, not smaller. You may be thinking of function limits, $\lim_{x\to a}f(x)$ where to get closer to the limit, you must get closer to a: smaller $\epsilon$ means smaller $\delta$. But with $\lim_{n\to \infty} a_n$ or even $\lim_{x\to\infty} f(x)$, your "a" is $\infty$ so you must get "closer to infinity" which means larger.

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