My problem from a text about limit of a sequence

[ShayanJ]

Hi
with having the sentence bellow:
«A number b is called the limit of a sequence x_{1}, x_{2}, . . . , x_{n}, . . . if for any \epsilon > 0 there is
N = N(\epsilon) such that |x_{n}– b| < \epsilon for all n > N.»
I have a question.does the N mean the number of sentence?
If yes could you explain me this example?
«Let us show that \lim_{n \rightarrow \infty}=\frac{n}{n+1}=1 .
Consider the difference |\frac{n}{n+1}-1|=\frac{1}{n+1} .The inequality \frac{1}{n+1}<\epsilon holds for all n>\frac{1}{\epsilon}-1=N(\epsilon) .
Therefore,for any positive \epsilon there is N=\frac{1}{\epsilon}-1 such that for n>N we have |\frac{n}{n+1}-1|<\epsilon .»
Because I solved the equation x_{n}=\frac{n}{n+1} due to n and I found n=\frac{x_{n}}{x_{n}+1} but N=\frac{1 - x_{n}}{x_{n}}
What is the problem?
And when I tried to use $ instead of [/tex] and ,In preview it didn&#039;t work and showed the exact code.<br /> And typing - caused «8211;».<br /> thanks

 

[uart]

Hi Shyan, what it means is this. Given any positive value for \epsilon, no matter how small, you can still find some large but finite number N such that all terms terms in the sequence after the Nth term differ from the limit b by less than this \epsilon. Does that make sense?

 

[ShayanJ]

Yes.In fact I understood that from the second time I read the paragraph but the example ruined everything by confusing me.I wanted someone to clear the example.

Because I solved the equation x_{n}=\frac{n}{n+1} due to n and I found n=\frac{x_{n}}{x_{n}+1} but N=\frac{1 - x_{n}}{x_{n}}

 

[uart]

Well in that case you've simply made some mistake in your algebra. Check again for you surely should get n = \frac{1-x_n}{x_n}

If you can't find your mistake then post your working and I (or someone else) will point it out to you.

Edit : Whoops. That should have been n = \frac{1-\epsilon}{\epsilon} there.

 

[ShayanJ]

I found my problem but my answer is still different just near to the text's answer.here's my calculation:
x_{n}=\frac{n}{n+1}\Rightarrow x_{n}(n+1)=n\Rightarrow x_{n}n+x_{n}=n\Rightarrow x_{n}n-n=-x_{n}\Rightarrow n(x_{n}-1)=-x_{n}\Rightarrow n=\frac{-x_{n}}{x_{n}-1}\Rightarrow n=\frac{x_{n}}{1-x_{n}}
I did the calculation three times.It seems the problem is really slight.

 

[Bleys]

Well, you have found n. You want to find N such that n>N. You can get
N=\frac{1 - x_{n}}{x_{n}} with a chain of inequalities and a bit of algebra.

I think it's valid if you just show that N>n, with the n you found and the N given to you, for all x_{n}. Unless ofcourse you're not given that N. All you have to note that is that x_{n} is positive and strictly less than 1 for all n (I'm guessing n is the natural numbers, but I don't think that matters since you want n large enough anyway).

 

[ShayanJ]

But if we say N&lt;n\Rightarrow N&lt;\frac{x_{n}}{1-x_{n}} we just get x_{n}&gt;\frac{1}{2}that is obviously useless.
And It is obvious that anyone who wants to check the correctness of his or her limiting action must be able to find N.So there must be a way for it.

 

[uart]

Yeah sorry Shyan that last post of mine was incorrect, I seem to have missed the fact that you were switching from an expression for "n" in terms of the error "epsilon" to an expression for "n" of terms of sequence value x_n.

So your last derivation was correct, n = \frac{x_n}{1-x_n}. But why do you want to calculate this quantity. You should follow the procedure in the first post and calculate "n" a function of the error (that is, the difference between x_n and 1). Anything else is irrelevant so just do it.