# Homework Help: Finding indexes.

1. Feb 12, 2010

### The_Iceflash

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

$$\left|\frac{n}{n+2}-1\right|< \epsilon$$

$$\left|\frac{-2}{n+2}\right| < \epsilon$$

$$\frac{2}{n+2} < \epsilon$$

$$\frac{n+2}{2} > \frac{1}{\epsilon}$$

$$n+2 > \frac{2}{\epsilon}$$

$$n > \frac{2}{\epsilon}-2$$

Last edited: Feb 12, 2010
2. Feb 12, 2010

### Staff: Mentor

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.

3. Feb 12, 2010

### The_Iceflash

The index:

I got n > 1998 as the index so,

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

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

4. Feb 12, 2010

### Staff: Mentor

This is to be expected in a sequence that is increasing, as {n/(n + 2)} is. A larger index gives you a larger value.

5. Feb 13, 2010

### HallsofIvy

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.