$a_1= \sqrt3 $ , all $n\geqq1$, $a_{n+1}=\sqrt{3+a_n}$, prove convergence of sequence ${a_n}$
How to prove... I don't know..
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Usual approach : Show that the sequence is bounded and increasing (or decreasing) – Peter Mar 23 '19 at 14:53
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For the limit $$\lim_{n\to \infty}a_{n}=\sqrt{3+a_n}\iff \lim_{n\to\infty}a_n=\frac{1+\sqrt{13}}{2}$$ – Dr. Mathva Mar 23 '19 at 15:07
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$(a_n)$ is bounded because $a_n<3$ for all $n\in\mathbb{N}$. You can show this by induction:
- $a_1<3$
- $a_n<3\implies a_{n+1}=\sqrt{3+a_n}<\sqrt{6}<3$
Also, $(a_n)$ is monotonically increasing:
- $a_{n+1}=\sqrt{3+a_n}>\sqrt{a_n+a_n}=\sqrt{2}\sqrt{a_n}>a_n$
We used that $a_n<3$ and thus $\sqrt{a_n}<\sqrt2$.
Because the series is monotonically increasing and bounded, it converges.
To find out the value $a$ it converges to, you have to solve $$a=\sqrt{3+a}\iff a^2=3+a\iff a=\frac12+\frac{\sqrt{13}}2\quad\text{because }a\geq 0.$$
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