Find the exact value of the limit of the sequence defined by $$a_1 = \sqrt 5, \;a_{n+1}= \sqrt {5 + a_n}$$
My only guess to this question is that the limit is e, however I don't know how to even begin to show this. Help please?
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3If $L$ is the limit (assuming one exists) then $L= \sqrt {5+L}\implies L^2-L-5=0$. – lulu Apr 19 '16 at 20:41
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2Special case of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ – YuiTo Cheng Jul 11 '19 at 08:10
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HINT: Suppose that the sequence actually does have a limit, and let that limit be $L$. Then
$$L=\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{5+a_n}=\sqrt{5+L}\;,$$
since the function $f(x)=\sqrt{5+x}$ is continuous. You can solve the equation $L=\sqrt{5+L}$ for $L$.
Corrected: It remains, however, to show that the sequence has a limit. Note that $a_{n+1}\ge a_n$ if and only if $\sqrt{5+a_n}\ge a_n$, which is true if and only if $5+a_n\ge a_n^2$, i.e., if and only if $a_n^2-a_n-5\le 0$. This is the case if and only if $a_n$ lies between the two roots of $x^2-x-5$. Try to show by induction on $n$ that this is always the case, so that the sequence, being bounded and increasing, must have a limit.
Brian M. Scott
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1@Bungo: You are entirely correct: I was sloppy. I’ve revised it accordingly. – Brian M. Scott Apr 19 '16 at 21:00