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Show and prove that the sequence $u_n = \frac{n^2+n-2}{4n^2-5}$ converges to $1/4$

Working:

To show the sequence converges to $1/4$, note that $\frac{n^2+n-2}{4n^2-5} = \frac{1+1/n -2/n^2}{4-5/n^2}$, and using the Algebraic Limit Theorem yields a limit of $1/4$.

The problem I am having is with the proof using the formal definition:

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I feel like my approach is way too long-winded, can someone please confirm whether it's correct and whether there's a quicker method? Anyhow, so I did:

We need to show for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that whenever $n \ge N$, it follows $|\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon$

Going backwards first: note that $|\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon \implies|\frac{4n-3}{4(4n^2-5)}| < \epsilon$. Now pick $n$ large enough so that $\frac{4n-3}{4(4n^2-5)} > 0$, hence $\frac{4n-3}{4(4n^2-5)} < \epsilon \implies 0 < n^2 - \frac{1}{4\epsilon}n + \frac{3}{16\epsilon}-\frac{5}{4}$

$\implies 0 < \left[n - \left(\frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon} \right) \right]\left[n - \left(\frac{1-\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon} \right) \right]$ so,

$n > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$ or $n < \frac{1-\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$.

Now pick $N > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$, then whenever $n \ge N > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$, we have $(8\epsilon n -1)^2 > 80\epsilon^2 - 12\epsilon + 1 \implies \epsilon > \frac{4n-3}{4(4n^2-5)} \implies |\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon$, as required.

I'm pretty sure I did the algebra right, just want to confirm whether all my arguments are correct and if there's a quicker method.


EDIT: Following on from @David's post, is this solution correct?

Need to show $\lim\left( \frac{n^2+n-2}{4n^2-5} \right) = 1/4$, note:

$|\frac{n^2+n-2}{4n^2-5} - \frac{1}{4} | = |\frac{4n-3}{4(4n^2-5)}|$. If $n \ge 2$, then $\frac{4n-3}{4(4n^2-5)} > 0$, so $|\frac{4n-3}{4(4n^2-5)}| = \frac{4n-3}{4(4n^2-5)}$. If $n^2 > 5$ (or more conveniently assume $n \ge 3$), then $\frac{4n-3}{4(4n^2-5)} < \frac{4n}{4(3n^2)} = \frac{1}{3n} < \epsilon$.

Now pick $N_1 \in \mathbb{N}$ such that $N_1 > \frac{1}{3\epsilon}$ and set $N = \max\{3, N_1\}$. Let $\epsilon > 0$ and suppose $n \ge N$, then $|\frac{n^2+n-2}{4n^2-5} - \frac{1}{4} | = |\frac{4n-3}{4(4n^2-5)}| = \frac{4n-3}{4(4n^2-5)} < \frac{4n}{4(3n^2)} = \frac{1}{3n} < \epsilon$, as required.

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It is nearly always far easier to simplify as much as you can before choosing $N$.

When I say "simplify", you don't need to simplify by finding $$\hbox{expression}=\hbox{simpler expression}.$$ You are trying to prove an inequality, not an equality, so it is good enough to find $$\hbox{expression}<\hbox{simpler expression}.$$ You have to be a bit careful though, as if the "simpler expression" is lots bigger than the "expression" it could change the whole problem. I would also recommend not saying "$<\varepsilon$" at the beginning but leaving it to the end.

There is a bit of a difficulty in your problem so first I will do a simpler one: show from the definition that $$\lim_{n\to\infty}\frac{n^2+n-2}{4n^2+5}=\frac{1}{4}\ .$$ In this case we could work as follows: assume that $n>N$, where $N$ is yet to be chosen. Then we have $$\eqalign{\left|\frac{n^2+n-2}{4n^2+5}-\frac{1}{4}\right| &=\left|\frac{4n-13}{4(4n^2+5)}\right|\cr &=\frac{4n-13}{4(4n^2+5)}\cr}$$ provided $n\ge4$. Now we can make a fraction bigger by increasing the numerator and/or decreasing the denominator. Since $4n-13<4n$ and $4n^2+5>4n^2$, we have $$\eqalign{\frac{4n-13}{4(4n^2+5)} &<\frac{4n}{4(4n^2)}\cr &=\frac{1}{4n}\cr &<\frac{1}{4N}\ ,\cr}$$ and this will be less than $\varepsilon$ provided $N>1/(4\varepsilon)$. So the proof would look something like this.

Let $\varepsilon>0$.
Choose $\displaystyle N=\max\Bigl(4,\,\frac{1}{4\varepsilon}\Bigr)$.
Suppose $n>N$. Then $$\left|\frac{n^2+n-2}{4n^2+5}-\frac{1}{4}\right|=\cdots\ \hbox{[working]}\ \cdots<\varepsilon\ .$$

I hope you will agree that this makes the algebra much easier than in your solution.

Your problem can be done in much the same way. The extra difficulty is that where we said $$4n^2+5>4n^2\ ,$$ you will now need to say $$4n^2-5>4n^2\ ,$$ and of course this is not true. Instead we can do something like this: if $n^2>5$ then $$4n^2-5>3n^2\ ;$$ now the working will go through much as before, though the details will be slightly different.

See how you go from here. Good luck!

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  • $\begingroup$ Thanks, this is brilliant, very concise and informative answer. I have edited my post for a solution using your explanations. Can you confirm if it is correct? Thanks. $\endgroup$
    – user40333
    Mar 22 '14 at 15:59
  • $\begingroup$ Looks good to me. I would just offer one comment on the setting out. Where you say for the first time $\frac{1}{3n}<\varepsilon$, you don't actually know this is true because you haven't yet specified $n$. Better to say at this stage something like "now we need $\frac{1}{3n}<\varepsilon$ when $n>N$, so choose $N$ as follows" etc. Then at the end you can say $\frac{1}{3n}<\varepsilon$ because you have taken $n$ to be greater than a suitably chosen $N$. $\endgroup$
    – David
    Mar 23 '14 at 0:10
  • $\begingroup$ Ah yes, good point, thank you! $\endgroup$
    – user40333
    Mar 23 '14 at 3:52
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You are to prove that $$\lim_{n\rightarrow\infty}\frac{4n-3}{4(4n^2-5)}=0$$

As you have said, the terms are positive for large $n$, so you can try to bound above the sequence with another simpler one: $$\frac{4n-3}{4(4n^2-5)}<\frac{4n}{4(4n^2-4n)}=\frac{1}{4(n-1)}$$

Any case, I think that your proof is correct (I have not verified the algebra).

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  • $\begingroup$ Ah great idea, so basically using the Squeeze Theorem :) $\endgroup$
    – user40333
    Mar 22 '14 at 12:12
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so i think other answer same : $n=\frac 1m$
$m\to 0$
$$\frac{\frac 1{m^2}+\frac 1m -2}{\frac 4{m^2}-5}= \frac{1+m-2\cdot m^2}{4-5\cdot m^2}$$

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  • $\begingroup$ please tell me about -1 vote $\endgroup$ Mar 22 '14 at 12:45

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