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 Post subject: Convergence of a seriesPosted: Mon Nov 09, 2015 5:16 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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: $\displaystyle \mathcal{S}=\sum_{n=1}^{\infty}a_n^r , \;\; r\in \mathbb{R}^+$.

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 Post subject: Re: Convergence of a seriesPosted: Wed Jul 26, 2017 4:58 am

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
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 $r \leq 2$

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