If a_n = sqrt{7+sqrt{7+sqrt{7+ldots}}} (n tim | vedclass

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(C) Let $a_{\infty} = \lim_{n 	o \infty} a_n$. Then $a_{\infty} = \sqrt{7 + a_{\infty}}$. Squaring both sides,we get $a_{\infty}^2 = 7 + a_{\infty}$,

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$a_n < 4, \forall n \geq 1$

Vedclass · Answer

(C) Let $a_{\infty} = \lim_{n 	o \infty} a_n$. Then $a_{\infty} = \sqrt{7 + a_{\infty}}$. Squaring both sides,we get $a_{\infty}^2 = 7 + a_{\infty}$,which implies $a_{\infty}^2 - a_{\infty} - 7 = 0$. Using the quadratic formula,$a_{\infty} = \frac{1 \pm \sqrt{1 - 4(1)(-7)}}{2} = \frac{1 \pm \sqrt{29}}{2}$. Since $a_n > 0$,we take the positive root: $a_{\infty} = \frac{1 + \sqrt{29}}{2} \approx \frac{1 + 5.385}{2} \approx 3.19$. Since the sequence $a_n$ is strictly increasing and bounded above by $a_{\infty}$,we have $a_n Thus,$a_n < 4$ is true for all $n \geq 1$.

If $a_n = \sqrt{7+\sqrt{7+\sqrt{7+\ldots}}}$ ($n$ times),then which one of the following is true?

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