Setting $a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\ $ I get that $$a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.$$ Clearly all numbers in the sequence are positive and we see that $a_n We can use the help-function $f(x)=\sqrt{2+x}$ such that $a_{n+1}=f(a_n).$ But since $$f'(x)=\frac{1}{2\sqrt{2+x}}=0\Leftrightarrow\text{no real solutions,}$$ $f(x)$ never flattens out or decreases, so it can't be convergent?
Sunday, 12 May 2019
calculus - Show that the sequence $sqrt{2}, sqrt{2+sqrt{2}}, sqrt{2+sqrt{2+sqrt{2+}}}...$ converges and find its limit.
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