Finding Indexes for \frac{n}{n+2} \approx 1 with varying epsilon values

[The_Iceflash]

I for the most part have this completed but I have a small question and thus checking if I did this correctly.

Homework Statement

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

Homework Equations

N/A

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<br /> <br /> <b>for any \epsilon > 0</b><br /> <br /> \left|\frac{n}{n+2}-1\right|< \epsilon<br /> <br /> \left|\frac{-2}{n+2}\right| < \epsilon<br /> <br /> \frac{2}{n+2} < \epsilon<br /> <br /> \frac{n+2}{2} > \frac{1}{\epsilon}<br /> <br /> n+2 > \frac{2}{\epsilon}<br /> <br /> n > \frac{2}{\epsilon}-2

 

[Mark44]

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.

 

[The_Iceflash]

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

 

[Mark44]

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

 

[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.