Easy divergence question made difficult

[kentt]

prove s_n=2n diverges

Show that for any L & for any e>0 and for and n* there exists an m>n* such that |s_n - L| >=e

I'm really a bit lost with this divergence. Especially frusterating as I've always just taken this as intuitively obvious. Thanks in advance for any help

-kentt

 

[daon]

Not sure what n* means, but here:

WTS: There exists a real number e>0 such that for all N>0, there exists an integer n>N such that |2n-L| > e.

First, show that S_n is strictly increasing.
S_{n+1} > S_{n} iff 2(n+1) > 2n iff 2n+2>2n iff 2>0. Done.

Now, show S_{n} > 0 for all n>0.
Base: n=1, S_{n}=2. Now, assume S_{n}>0, then S_{n+1} = 2(n+1) > 2n = S_n > 0. Done.

The above also gives us that if such an L were to exist, it must be positive and greater than any s_{n} for all n. Otherwise L would be less than S_n for some n and since S_n is strictly positive and increasing, S_n would never return to L.

So, let e=1, let N>0, and take n=2N>N. then |2n-L| = |4N-L| = |2N + (2N-L)| > |2N|, since L>2N implies 2N-L < 0.

And |2N| = 2N > 1 = e.

The above inequality comes from N>0 which implies N > 1 which implies S_N = 2N > 2 > 1=e.

-Daon

 

[kentt]

Thank you kindly. That's a very good explination. Hopefully with that in mind I will be able to finish of my homework.

 

[pka]

Daon what you have done is o.k.
But the question requires that it be done for any L & for any e>0.
Have you done that?

When you say "if such an L were to exist" what does that mean? I do not see an 'L' in the proof before that.

 

[daon]

Daon what you have done is o.k.
But the question requires that it be done for any L & for any e>0.
Have you done that?

When you say "if such an L were to exist" what does that mean? I do not see an 'L' in the proof before that.

Sorry, I was under the impression the problem was to show divergence of S_n.

By "if such an L were to exist" I meant that if S_n=2n had a limit, it must be positive and greater-than or equal to s_n for any n because it is strictly increasing. I didn't mean this as part of the "proof", just an explanation for the OP.

Its quite possible I may have misstepped in my reasoning. I was basically saying that for any L it must be true, but that the only Ls that make sense must be greater-than or equal to S_n for any n. Might this proof have been better done by contradiction?

 

[pka]

I was under the impression the problem was to show divergence of S_n.

Yes that is exactly right. And that is what you were doing with your proof.
However, it seems that the poster had a particular set of steps to prove that.
The way most of us would prove it is what you did: if L exists then that would imply that the positive even integers are bounded above.

 

[pka]

While writing a lecture for quite a different purpose it dawned on me what any why this problem required those steps. And now I think it is a very good approach.

Recall that \(\displaystyle \left| {\left| x \right| - \left| y \right|} \right| \le \left| {x - y} \right|.\)
Given any \(\displaystyle L\quad \& \quad \varepsilon > 0\) using the ARCHIMEDEAN PROPERTY we get \(\displaystyle \left( {\exists K \in Z^ + } \right)\left[ {K > \frac{{\left| L \right| + \varepsilon }}{2}} \right].\)
Now if \(\displaystyle \begin{array}{rcl}
m \ge K\quad & \Rightarrow & \quad 2m > \left| L \right| + \varepsilon \\
& \Rightarrow & S_m > \left| L \right| + \varepsilon \\
& \Rightarrow & S_m - \left| L \right| > \varepsilon \\
& \Rightarrow & \left| {S_m - L} \right| > \left| {S_m - \left| L \right|} \right| > \varepsilon \\ \end{array}.\)