I just dont get this proof of convergence of even and uneven sequence coeffi. ?

[hsmathfreaker]

Homework Statement

Show that if Lim(n-->inf.)(a_(2n)-->L) and Lim(n-->inf.)(a_(2n+1)-->L) then Lim(n-->inf.)(a_n-->L).

The Attempt at a Solution

I just don't get this; I can see the big picture though. If the odd coeffictions of a sequences goes towards one the same number as the even coefficients, then ultimately the complete sequence must also be approaching that number.

But the proof in my textbooks uses the formal limit defintion of the limit of a sequence. It states the the odd part has an N_1 and the even part has N_2 and so on .. you guys know the definition:) But then it goes on into something about a "max" of those two numbers, which i have abselutely no idea what is?? And looking at it, i cannot seem to understand it .. Its really annoying, since all the assignemts were going well. I thought i had these simple infinite series, but now i am completely blank lol .. I have no idea how to formulate a proof of my understanding.

It may be good to mention, that i am a 17 year old high school student on my own, self designes study course, since the HS math is rather boring^^ Maybe i am missing a key component in my mathematical knowledge to do this proof, or i may simply not be smart enough ..

Could you guys elaborate a bit ? :) I'm sorry i haven't provided an attempt, but as i said, I am rather blank ..

 

[mighty2000]

First of all, don't be discouraged by the fact that you are only 17 years old and studying on your own. It takes a lot of dedication and motivation to do so and you should be proud of yourself for taking on this challenge. As for your question, let me try to explain the proof in a simpler way.

To prove that Lim(n-->inf.)(a_n-->L), we need to show that for any given epsilon > 0, there exists a natural number N such that for all n > N, |a_n - L| < epsilon.

Now, since Lim(n-->inf.)(a_(2n)-->L), we know that there exists a natural number N_1 such that for all n > N_1, |a_(2n) - L| < epsilon. Similarly, since Lim(n-->inf.)(a_(2n+1)-->L), there exists a natural number N_2 such that for all n > N_2, |a_(2n+1) - L| < epsilon.

Now, let N = max{N_1, N_2}. This means that for all n > N, both |a_(2n) - L| < epsilon and |a_(2n+1) - L| < epsilon.

But since a_n is just a combination of a_(2n) and a_(2n+1), we can write a_n as a_n = (a_(2n) + a_(2n+1)) / 2. Using this, we can show that |a_n - L| < epsilon for all n > N, which proves that Lim(n-->inf.)(a_n-->L).

I hope this helps and that you now have a better understanding of the proof. Keep up the good work and don't hesitate to ask for help when needed. Good luck with your studies!