# Limit of $a_n = -\cos \left( \frac{\pi}{8n-2} \right)$ (proof)

Warning to anyone who stumbles upon this: It is wrong completely and utterly don't use it for reference, thank you Don and Gerry for helping me see this

So my first question is asking whether or not the following is a valid proof:

Claim: $a_n = -\cos \left( \frac{\pi}{8n-2} \right) \to 1 \text{ as } n \to \infty$

Given $\epsilon > 0$ let $N = \Big[ \frac{\pi}{8 \arccos \left( \epsilon - 1 \right) } + \frac{1}{4} \Big] + 1 \; \forall \; 0 \le \epsilon \le 2$ so then it is true that

$$\forall \; n \ge N = \Big[ \frac{\pi}{8 \arccos \left( \epsilon - 1 \right) } + \frac{1}{4} \Big] + 1 > \frac{\pi}{8 \arccos \left( \epsilon - 1 \right) } + \frac{1}{4} \\ \implies 8n > \frac{\pi}{\arccos \left( \epsilon - 1 \right)} + 2 \\ \implies 8n - 2 > \frac{\pi}{\arccos \left( \epsilon - 1 \right)} \\ \implies \arccos \left( \epsilon - 1 \right) > \frac{\pi}{8n-2} = \Bigl\lvert \frac{\pi}{8n-2} \Bigr\rvert \\ \implies \epsilon - 1 > \cos \left( \Bigl\lvert \frac{\pi}{8n-2} \Bigl\lvert \right) = \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert \\\\ \implies \epsilon > \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert + 1 = \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert + \lvert 1 \rvert \ge \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) + 1 \Bigr\rvert \\\\ \iff \epsilon > \Bigl\lvert \, - \cos \left( \frac{\pi}{8n-2} \right) - 1 \Bigr\rvert$$ and then it's trivial to see that the above still holds for $\epsilon > 2$

$\therefore \; a_n \to 1$

The second question is silly since the first part is invalid...

My next question is how come you cannot replace 1 with -1 throughout this proof to show that the limit converges to -1. I'm pretty sure it goes to 1 so that's why I picked 1 but wouldn't replacing the 1 with -1 simply change the absolute value to be from | 1 | to the |- 1| where both are equivalent to 1?

In other words why is this wrong?

$$\forall \; n \ge N = \Big[ \frac{\pi}{8 \arccos \left( \epsilon - 1 \right) } + \frac{1}{4} \Big] + 1 > \frac{\pi}{8 \arccos \left( \epsilon - 1 \right) } + \frac{1}{4} \\ \implies 8n > \frac{\pi}{\arccos \left( \epsilon - 1 \right)} + 2 \\ \implies 8n - 2 > \frac{\pi}{\arccos \left( \epsilon - 1 \right)} \\ \implies \arccos \left( \epsilon - 1 \right) > \frac{\pi}{8n-2} = \Bigl\lvert \frac{\pi}{8n-2} \Bigr\rvert \\ \implies \epsilon - 1 > \cos \left( \Bigl\lvert \frac{\pi}{8n-2} \Bigl\lvert \right) = \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert \\\\ \implies \epsilon > \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert + 1 = \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) \Bigr\rvert + \lvert -1 \rvert \ge \Bigl\lvert \, \cos \left( \frac{\pi}{8n-2} \right) - 1 \Bigr\rvert \\\\ \iff \epsilon > \Bigl\lvert \, - \cos \left( \frac{\pi}{8n-2} \right) + 1 \Bigr\rvert$$ which would then lead me to $a_n \to -1$ rather than 1

• Presumably, all these limits are as $n\to\infty$. If so, please edit this into the question. – Gerry Myerson Mar 11 '13 at 4:39
• Hollie Mollie, some calculations! Say: can't you use known facts about continuous functions? – DonAntonio Mar 11 '13 at 4:40
• Cosine is a decreasing function, so $\arccos a\lt b$ implies $a\gt\cos b$ – Gerry Myerson Mar 11 '13 at 4:43
• @DonAntonio I don't think I follow but I'm guessing not (I'm learning fairly restrictively, pretending we don't know what a function is yet, only sequences) – DanZimm Mar 11 '13 at 4:43
• @GerryMyerson Yep, fixed in above I apologize – DanZimm Mar 11 '13 at 4:43

At request of OP: Cosine is a decreasing function, so $\arccos a\lt b$ implies $a\gt\cos b$.
$$\lim_{n\to\infty}-\cos\left(\frac{\pi}{8n-2}\right)=-\cos\left(\lim_{n\to\infty}\frac{\pi}{8n-2}\right)=-\cos (0)=-1$$
$\cos(2x) = \cos^2(x) - \sin^2(x) = 1-2\sin^2(x)$ so $1-\cos(x) = 2\sin^2(x/2)$.
If we can use $|\sin(x)| \le |x|$ then $|1-\cos(x)| \le 2|x/2|^2 = |x|^2/2$ so $|1-\cos\left (\frac{\pi}{8n-2}\right )| \le \left ( \frac{\pi^2}{2(8n-2)^2}\right )$ and the rest should be easy.