Proving Convergence and Polarization Identity for Limiting Sequences

[mynameisfunk]

Homework Statement

I have a 3 parter. I think I have the first part, the second part I am not sure of and the third part I am kind of getting desperate with...

(a) Prove first that if c is a constant, and \lim_{n\rightarrow \inf} s_n = s , then \lim_{n\rightarrow \inf} cs_n = cs.
(b) Prove that if \lim_{n\rightarrow \inf} s_n = s then \lim_{n\rightarrow \inf} s^2_n = s^2. (Hint:recall that convergent sequences are bounded.)
(c) Use the polarization identity ab = \frac{1}{4}((a+b)^2-(a-b)^2) to prove that if \lim_{n\rightarrow \inf} s_n = s and \lim_{n\rightarrow \inf} t_n = t , then \lim_{n\rightarrow \inf} t_ns_n = ts.

Homework Equations

The Attempt at a Solution

(a) Since for all \varepsilon > 0 , there exists an N, such that if n > N , |s_n-s| < \varepsilon , there should
exist an n_{\mu} \geq n such that c|s_n-s| \leq \varepsilon hence \lim_{n\rightarrow \inf} cs_n = cs

(b) \lim_{n \rightarrow \inf}s_n = s implies that {s_n} is compact and bounded. Take \varepsilon > 0. For some M, if n \geq M, then |s_n-s|=|s-s_n| < \sqrt{\varepsilon} and therefore |s-s_n|^2 < \varepsilon. so now \lim_{n \rightarrow \inf}(s_n-s)^2 = 0.
Take |s_n^2-s^2| = |s_n(s_n-s)-s(s-s_n)| \leq |s_n||s_n-s|-|s||s-s_n| \leq |s_n-s|(|s_n|-|s|) \leq \varepsilon which gives us \lim_{n \rightarrow \inf}(s_n^2-s^2)=0

(c) If my other two are even correct, this is where I am stuck. PLEASE give me a little nudge...

 

[╔(σ_σ)╝]

1) It would be nice if you showed more steps for your choice of epsilon but the proof is more or less correct.
2) I can't say the proof is incorrect but you should have followed the suggestion. The proof is kind of ...hmm... I don't know how to put it.:-)

s_{n}^2 - s^2 = \left(s_n+s \right) \left( s_n -s \right)

You know abs(s_n) <=B . You should be able to find a better proof.

3) I am not sure what you are allowed to use but it could be done using some limit theorems; I have never "seen" that inequality. I suppose if you plug in s_nt_n in that forumula and use some other limit theorem you could do this problem. The formula looks like some sort of completing the square formula ... Anyway
(s_n)(t_n) -st = (t_n)( s_n -s) + s(t_n-t)

Use trigangle inequality and the fact that t_n is bounded and then use the typical epsilon/2 argument with a bit of modification.

 

[ZioX]

for (c) you would use (a) and (b)

 

[mynameisfunk]

I am still having a lot of trouble with this. i know it must be obvious.
I took the upper bound of t_n = B i went and plugged in t_ns_n to get t_ns_n=\frac{1}{4}(((s_n+t_n)^2-(s_n-t_n)^2) \leq \frac{1}{4}(s_n^2+2Bs_n+B^2-s_n^2+2s_nB) = s_nB which converges to sB and then the same argument but take an upper bound of s_n = A and so on. Then from (a), we have in both cases that A and B are constants and they both converge to Bs and At, respectively. But I feel like i did not really use the formula, wasn't it just basically a formality? I could have skipped that part and I wouldn't see anything wrong with it. Am i doing this right?

 

[╔(σ_σ)╝]

That is why I suggested another method.

With the formula you seem to need some other results like ...the sum of convergent sequences is convergent .

 

[ZioX]

<br /> a=t_ns_n-st \quad b=t_ms_m-st<br />

and think of Cauchy sequences.

 

[mynameisfunk]

ZioX, what is the relation of t_n to t_m?

 

[ZioX]

Same sequence, just at different n.

Another way of looking at whether sequences converge is that after some point they become arbitrarily close together. That is, if for any epsilon there is an N such that for n,m > N |x_n-x_m|< epsilon.

Such a sequence is called Cauchy sequence.

 

[╔(σ_σ)╝]

Why is it so difficult to do the following \left(s_n \right) \left(t_n \right) -st = \left(t_n \right) \left( s_n -s \right) + s \left(t_n-t \right)

\left|\left(t_n \right) \left( s_n -s \right) + s \left(t_n-t \right)\right| \leq \left|t_n\right|\left|s_n -s\right| + s\left|t_n -t\right|We know \left|t_n -t\right| \leq \frac{\epsilon}{s} for some n_0.

We know \left| t_n \right| \leq B

And we also know ...
\left|s_n -s\right| \leq \frac{\epsilon}{B} for some n_1