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Question 03 (a)
Use the definition of convergence for sequences, prove that
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
A sequence converges to a number L if for any there exists a number such that
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
This is the proof as I worked it out; Solution 2 presents it as you can polish it and write it nicely to impress your teacher or nerdy friends.
I fix myself an , that means, it's as if I have a fixed value in my mind like . I need to show that if is large enough (that is, larger than some number I call ) then I will have that the term is -close to my number L (which here I'm already told should be 1). In other words, if is large enough, then I should have
Here I know the sequence really well, I'm told that
So I can rewrite the not-so-good-looking term into something much more presentable by using a little algebra:
And since is a positive integer, the above fraction is clearly always positive so the absolute value is here totally useless. So I'll rewrite
Now it seems indeed that I can make this as small as I want as long as is large enough. I still need to be precise about this and explicitly guarantee how big has to be for the term to be less than . Here are two ways to do this:
First way: I can directly solve
It gives me
which I can rewrite as
Now since is some fixed number, the above is just a quadratic equation (something like with an ugly coefficient in front of the term). And since it corresponds to a positive parabola (I like to think them as smiling parabolas) I know for sure that it will be positive (above the x-axis) for any value larger than its second x-intercept (if there are any x-intercepts). The x-intercepts are the solutions of the equation so using the quadratic formula I find
The larger solution has to be the one using the "+" sign, so I get that
meaning that any integer greater than or equal to works as . You can verify this for yourself; if you choose , then , and for any value of greater than or equal to , you can see that .
Second way: As is often the case in math, there is another way that is easier, but that involves being a little bit more smart about all this. Since I don't care that the value of is the best possible, just that it works, I can do the following manipulation
And now since is some positive integer, it is always true that and therefore that
And so we get something much simpler to deal with:
Now we just determine when it is true that
Clearly, this occurs when
So any integer greater than or equal to can be used to establish convergence of the sequence. You can again verify this for yourself; if you choose , then you will get here , which we know from the above work is more than large enough to guarantee that for any value of greater than or equal to that, we will have . This is much simpler; we only lost some precision on how big the number has to be, but to prove convergence, we really don't care, so we might as well make our lives easier.
Note: this was the work done as you go, which corresponds to how you would find out how to prove that the given sequence converges. Now that we have done that work, we can present our work concisely. To see that, check out Solution 2.
This is the polished proof that we obtained in Solution 1 and resembles more what mathematicians present in books.
Fix a value of and let be so large that
Then, if ,
which concludes our proof.