Science:Math Exam Resources/Courses/MATH220/April 2011/Question 03 (a)

MATH220 April 2011

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Question 03 (a)
Use the definition of convergence for sequences, prove that $\lim _{n\to \infty }{\frac {n^{2}+3n+1}{n^{2}}}=1.$

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!

Hint
A sequence ${a_{n}}$ converges to a number L if for any $\epsilon >0$ there exists a number $N_{\epsilon }$ such that ${\text{if }}n\geq N_{\epsilon }\quad {\text{then}}\quad |a_{n}-L|<\epsilon$

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\epsilon >0$, that means, it's as if I have a fixed value in my mind like $\epsilon =0.01$. I need to show that if $n$ is large enough (that is, larger than some number I call $N_{\epsilon }$) then I will have that the term $a_{n}$ is $\epsilon$-close to my number L (which here I'm already told should be 1). In other words, if $n$ is large enough, then I should have $\displaystyle |a_{n}-1|<\epsilon$ Here I know the sequence really well, I'm told that $a_{n}={\frac {n^{2}+3n+1}{n^{2}}}$ So I can rewrite the not-so-good-looking term $|a_{n}-1|$ into something much more presentable by using a little algebra: ${\begin{aligned}|a_{n}-1|&=\left|{\frac {n^{2}+3n+1}{n^{2}}}-1\right|\\&=\left|{\frac {n^{2}+3n+1}{n^{2}}}-{\frac {n^{2}}{n^{2}}}\right|\\&=\left|{\frac {n^{2}+3n+1-n^{2}}{n^{2}}}\right|\\&=\left|{\frac {3n+1}{n^{2}}}\right|\\\end{aligned}}$ And since $n$ is a positive integer, the above fraction is clearly always positive so the absolute value is here totally useless. So I'll rewrite $|a_{n}-1|={\frac {3n+1}{n^{2}}}$ Now it seems indeed that I can make this as small as I want as long as $n$ is large enough. I still need to be precise about this and explicitly guarantee how big $n$ has to be for the term $|a_{n}-1|$ to be less than $\epsilon$. Here are two ways to do this: First way: I can directly solve ${\frac {3n+1}{n^{2}}}<\epsilon$ It gives me $\displaystyle 3n+1<\epsilon n^{2}$ which I can rewrite as $\displaystyle 0<\epsilon n^{2}-3n-1$ Now since $\epsilon$ is some fixed number, the above is just a quadratic equation (something like $\epsilon x^{2}-3x-1$ with an ugly coefficient in front of the $x^{2}$ 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 $\epsilon x^{2}-3x-1=0$ so using the quadratic formula I find $x={\frac {3\pm {\sqrt {9+4\epsilon }}}{2\epsilon }}$ The larger solution has to be the one using the "+" sign, so I get that $0<\epsilon n^{2}-3n-1\quad {\text{if}}\quad n>{\frac {3+{\sqrt {9+4\epsilon }}}{2\epsilon }},$ meaning that any integer greater than or equal to ${\frac {3+{\sqrt {9+4\epsilon }}}{2\epsilon }}$ works as $N_{\epsilon }$. You can verify this for yourself; if you choose $\epsilon =0.01$, then ${\frac {3+{\sqrt {9+4\epsilon }}}{2\epsilon }}\approx 300.33$, and for any value of $n$ greater than or equal to $N_{\epsilon }=301$, you can see that $(3n+1)/n^{2}<0.01$. 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 $N_{\epsilon }$ is the best possible, just that it works, I can do the following manipulation ${\frac {3n+1}{n^{2}}}={\frac {3}{n}}+{\frac {1}{n^{2}}}$ And now since $n$ is some positive integer, it is always true that $n\leq n^{2}$ and therefore that ${\frac {1}{n^{2}}}\leq {\frac {1}{n}}.$ Hence ${\frac {3}{n}}+{\frac {1}{n^{2}}}\leq {\frac {3}{n}}+{\frac {1}{n}}={\frac {4}{n}}.$ And so we get something much simpler to deal with: $|a_{n}-1|={\frac {3n+1}{n^{2}}}\leq {\frac {4}{n}}$ Now we just determine when it is true that ${\frac {4}{n}}<\epsilon .$ Clearly, this occurs when ${\frac {4}{\epsilon }}<n.$ So any integer $N_{\epsilon }$ greater than or equal to $4/\epsilon$ can be used to establish convergence of the sequence. You can again verify this for yourself; if you choose $\epsilon =0.01$, then you will get here ${\frac {4}{\epsilon }}=400$, which we know from the above work is more than large enough to guarantee that for any value of $n$ greater than or equal to that, we will have $|a_{n}-1|<0.01$. This is much simpler; we only lost some precision on how big the number $N_{\epsilon }$ 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.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This is the polished proof that we obtained in Solution 1 and resembles more what mathematicians present in books. Fix a value of $\epsilon >0$ and let $N_{\epsilon }\in \mathbb {N}$ be so large that $N_{\epsilon }>{\frac {4}{\epsilon }}.$ Then, if $n\geq N_{\epsilon }$, ${\begin{aligned}|a_{n}-1|&=\left|{\frac {n^{2}+3n+1}{n^{2}}}-1\right|\\&={\frac {3n+1}{n^{2}}}\\&={\frac {3}{n}}+{\frac {1}{n^{2}}}\\&\leq {\frac {3}{n}}+{\frac {1}{n}}\\&={\frac {4}{n}}\\&\leq {\frac {4}{N_{\epsilon }}}\\&<\epsilon \end{aligned}}$ which concludes our proof.

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