the limits of $a_n $and $b_n$

Question

this is related to that one $a_n$ is bounded and decreasing

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 $c_{n}=a_{n}\sin\left(\dfrac{\pi }{2^{n}}\right)$

Deduce the value of the limits of $a_n, b_n$, and $c_n$.

You literally linked to the answer. – Shahar Dec 24 '14 at 21:00 — Shahar, Dec 24 '14 at 21:00

score 2 · Answer 1 · answered Dec 24 '14 at 20:19

2

Hint

Notice that

$$\cos\left(\frac \pi{2^k}\right)=\frac12\frac{\sin\left(\frac \pi{2^{k-1}}\right)}{\sin \left(\frac \pi{2^k}\right)}$$ and telescope.

answered Dec 24 '14 at 20:19

Thanks but i was working from the morning i can't take hint if possible details – Educ Dec 24 '14 at 20:21
2

So leave it for tomorrow;-) – Dec 24 '14 at 20:23
but i need it now – Educ Dec 24 '14 at 21:00

score 0 · Answer 2 · answered Dec 24 '14 at 20:25

0

From a related post and its solution, $a_n \to \dfrac{2}{\pi} \Rightarrow b_n \to \dfrac{2}{\pi}$ as well. Plus $c_n \to 0$.

answered Dec 24 '14 at 20:25

DeepSea

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the limits of $a_n $and $b_n$

2 Answers2

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