$0\leq a_{n}-l\leq \dfrac{\pi^{2} }{2^{2n+1}}$

Question

this is related to that one the limits of $a_n $and $b_n$

Let for $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$ and let $c_{n}=a_{n}\sin\left(\dfrac{\pi }{2^{n}}\right)$

Show that $0\leq a_{n}-l\leq \dfrac{\pi^{2} }{2^{2n+1}}$ (note that: $|1-\cos x|\leq \dfrac{x^{2}}{2})$

can we say that $$b_n < l < a_n$$
Then $$ 0 < a_n-l< a_n-b_n$$

$$a_n- b_n = a_n(1-\cos(\dfrac{\pi}{2^n})$$

since $|a_n|\leq 1$ and $1-\cos(\dfrac{\pi}{2^n})\leq \dfrac{\pi}{2^{n+1}} $

am i right

I think its right......but you need to square the $\pi$, and double the $n$. — DeepSea, Dec 24 '14 at 21:20
but i need to add something to make me sure to say that we've $b_n < l < a_n$ — Educ, Dec 24 '14 at 21:24

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

2

As shown in this answer, $$ a_n=\frac2{2^n\sin\left(\frac\pi{2^n}\right)} $$ therefore $$ b_n=\frac2{2^n\tan\left(\frac\pi{2^n}\right)} $$ It is also shown that $$ l=\frac2\pi $$

Furthermore, as shown in this answer, for $|x|\lt\frac\pi2$, $$ \frac{\sin(x)}{x}\le1\le\frac{\tan(x)}{x} $$ which implies that $$ b_n\le\frac2\pi\le a_n $$ and that $$ 1-\cos(x)\le\frac{x^2}2 $$ therefore $$ 1-\cos\left(\frac\pi{2^n}\right)\le\frac{\pi^2}{2^{2n+1}} $$ So you are correct, but there are better bounds for $1-\cos(x)$.

edited Apr 13 '17 at 12:20

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answered Dec 24 '14 at 21:35

robjohn

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thanks so much just give me moment to read what you wrote – Educ Dec 24 '14 at 21:44

$0\leq a_{n}-l\leq \dfrac{\pi^{2} }{2^{2n+1}}$

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