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Convergence of a series

Posted: Mon Nov 09, 2015 5:16 pm
by Tolaso J Kos
Let \( a_n =\underbrace{\sin \left ( \sin \left ( \sin \cdots (\sin x) \cdots \right ) \right )}_{n \; \rm {times}} \) and \( x \in (0, \pi/2) \). Examine if the series:

$$ \mathcal{S}=\sum_{n=1}^{\infty} a_n $$

converges.

Do the same question for the series: \( \displaystyle \mathcal{S}=\sum_{n=1}^{\infty}a_n^r , \;\; r\in \mathbb{R}^+ \).

Re: Convergence of a series

Posted: Wed Jul 26, 2017 4:58 am
by dr.tasos
First $ a_{n+1}=sin(a_n) \leq a_{n} $
Since $ a_1 \in (0, \frac{\pi}{2} ) $ then using induction one can prove $ a_n \in (0, \frac{\pi}{2}) $

Hence $ a_n $ being bounded from below $ \lim_{n \to \infty} a_n=0 $
To estimate what the sequence feels like we compute the following limit

$$ \lim_{n \to \infty} n a_n^2 $$

But to use cezaro stolz lemma i will compute this one instead :
$ \lim_{n \to \infty} \frac{1}{n a_n^2}=\lim_{n \to \infty} \frac{a_n^2-a_{n+1}^2}{a_{n+1}^2a_n} $

So in order to compute that we compute it's "continuous" version and then we jump on the discrete case using continuity .

$ \lim_{x \to 0} \frac{x-sin^2x}{xsin^2(x)} $ dividing both sides with $ x^3 $ we and after using DLH rule several times we get that $ \lim_{x \to 0} \frac{x-sin^2x}{xsin^2(x)}=\frac{1}{3} $

Hence $ \lim_{n \to \infty}\frac{1}{n a_n^2} =\frac{1}{3} \Rightarrow \lim_{n \to \infty} n a_n^2 = 3 \Rightarrow \lim_{n \to \infty} \sqrt{n} a_n=\sqrt3 $

So using comparison test with $ \frac{1}{\sqrt{n}} $ our series diverges .

For the second part comparing with $ (\frac{1}{\sqrt{n}})^r $ we get that it converges if and only if $ 2r>4 \Leftrightarrow r>2 $ and diverges if $ r \leq 2 $