Question

I need help taking out a variable. The first pdf file is the solution to a similar problem that I'm working out (number 16.6). In this problem, I need to take out 'n' from the absolute value. The actual problem I am working on is in the second pdf file number 1a) Could someone help me solve the inequality for n?

Answer 1

which one?

Answer 2

the second pdf file has the problem i am working on right now, num 1a)

Answer 3

Do you know the definition of sequence convergence?

Answer 4

a sequence Sn is said to converge to the real number s provided that for each E > 0 there exists a real number N such that for all n (member of) Natural number, n > N implies that |Sn - s| < E.

Answer 5

If Sn converges to s then s is called the limit of the sequence Sn, if a sequence doesn't converge to a real number it is said to diverge.

Answer 6

The solution to similar problem is in the first pdf attachment number 16.6

Answer 7

I got stuck at the part where they take out the variable 'n' to get (according to the first pdf number 16.6 a) n > |k|/E

Answer 8

oh this is just going back to delta epsilon stuff =)

Answer 9

The problem i am working on (second pdf number 1a) , I don't know how to take out 'n' from the problem so I can get the inequality n > something

Answer 10

and yes more delta epsilon :)

Answer 11

They give us epsilon, and we give them N

Answer 12

yes

Answer 13

what did you have so far?

Answer 14

i got up to | n \ (n^2 +4n+4)| ( i just squared both numerator and denominator)

Answer 15

i know, haven't got far :/

Answer 16

Since we know that \(s=0\) and \(n>0\) we can get rid of those absolute values pretty instantly.

Answer 17

oh ok

Answer 18

Now, we can use something like squeeze theorem here...

Answer 19

But we don't really need to squeeze from the bottom, only from the top, and we don't actually need to have a tight squeeze either.

Answer 20

\[
\frac{n }{n^2 +4n+4} = \frac{1}{n+4+\frac{1}{n}}
\]

Answer 21

oh ok!

Answer 22

I'm trying to think here....not finished yet...

Answer 23

now it's clear that: \[
\frac{1}{n+4+\frac{1}{n}} < \frac{1}{n} < \epsilon
\]

Answer 24

So I think \[
N = \left\lceil \frac{1}{\epsilon}\right\rceil

\]would be a good rule.... what do you think?

Answer 25

If they want an accuracy of \[
\epsilon = 0.0001
\]We give them \[
N = \left\lceil \frac{1}{\epsilon}\right\rceil = 10,000
\]

Answer 26

Whoops, we squared both sides... I forgot!! It should be \(\epsilon^2\)

Answer 27

hmm so i should always in cases like this, relate like this: fraction i have < fraction that is easier to image and use?

Answer 28

oh wait but why does it matter if we squared it? does it make a difference on epsilon?

Answer 29

Because remember when you squared both sides? You had to square epsilon too.

Answer 30

oh ok, but if we multiply both sides by a/n (squeeze thm) we don't have to do that to epsilon?

Answer 31

\[
\frac{\sqrt{n}}{n+2}< \epsilon\implies \left(\frac{\sqrt{n}}{n+2}\right)^2<\epsilon^2
\]

Answer 32

When I said squeeze theorem, I was being very unprofessional and thinking aloud.

Answer 33

The point is we want to find an upper bound for our function which would be easy to find the inverse of.

Answer 34

For example, the inverse of \[
\frac{1}{n+4+\frac{1}{n}}
\]Is really crazy

Answer 35

ohh ok that makes sense!

Answer 36

But the inverse of \(1/n\) is just \(1/n\)

Answer 37

oh ok, thank you!! ^_^

Answer 38

Do you want to try 1b now?

Answer 39

kk!

Answer 40

having trouble?

Answer 41

so i got upto (18+5n)/(9n^2-3n)
since (18+5n) < (9n^2-3n), can we say that it is simlar to 1 / (9n^2-3n)

Answer 42

how did you get there?

Answer 43

or like 16.6 e) we can say that for n > 3, 18 +5n < 5n

Answer 44

actually i think my first response is incorrect, i think i should stick to the response that i just did above

Answer 45

i made a mistake:
for n > 3, 18 +5n < 10n

Answer 46

What did you get for: \[
\frac{6+5n^2}{2n^2-n} - \frac{3}{5}
\]

Answer 47

I mean \[
\frac{6+5n^2}{2n^2-n} - \frac{5}{3}
\]

Answer 48

i got |(18+5n)/(9n^2-3n) | and then i removed abs value signs b:c for sure it is positive

Answer 49

Hmmm

Answer 50

Divide the whole thing by \(n\) and look for a function which bounds it above, and would be easy to get the inverse of.

Answer 51

so \[(\frac{ 18 }{ n } + 5) / (9n-\frac{ 3 }{ n })\] < 5/9n

Answer 52

for n > 4

Answer 53

So then, what do you get when you solve for \(\epsilon\) in terms of \(n\)?

Answer 54

n > \[\frac{ 5 }{ 9E}\]

Answer 55

then N = max {5/9E , 5}

Answer 56

Looks great.

Answer 57

these are much easier normal limits because you don't have to worry about absolute value bars.

Answer 58

yeah!!

Answer 59

thank you so much for helping me!!!