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On the sequence $x_{n+1} = \sqrt{c+x_n}$

Where does this sequence converge? $\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...

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    $\begingroup$ @Marvis, it's a little weird to close a question from November 2010 as a duplicate of a question from March 2012, isn't it? $\endgroup$ Jun 20, 2012 at 7:13
  • $\begingroup$ @GerryMyerson The other question was repurposed to cut down on duplicates. $\endgroup$
    – user17762
    Jun 20, 2012 at 7:14
  • $\begingroup$ @Marvis, maybe this question should have been repurposed, instead? Maybe the general issue should be taken up on meta, "retroactive closure". $\endgroup$ Jun 20, 2012 at 7:35
  • $\begingroup$ @GerryMyerson Yes. I too think so. I also believe that there should be an option to merge questions and corresponding answers. This thread (meta.math.stackexchange.com/questions/3746/…) is probably the right thread to voice our opinion on this issue. $\endgroup$
    – user17762
    Jun 20, 2012 at 7:54
  • $\begingroup$ If it was 'a' instead of 7 there, then the answer would be (1+sqrt(1+4*a))/2. Sometimes pattern in these problems start after a few entries,you need to just find the pattern and solve it separately and put it back into the the problem. $\endgroup$
    – Aang
    Jun 20, 2012 at 10:11

4 Answers 4


For a proof of convergence,

Define the sequence as

$\displaystyle x_{0} = 0$

$\displaystyle x_{n+1} =\sqrt{7 + x_n}$

Note that $\displaystyle x_n \geq 0 \ \ \forall n$.

Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.

We claim the following:

i) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \lt b$
ii) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \gt x_n$

For a proof of i)

We have that

$\displaystyle x_n \lt b = b^2 - 7$ and so $x_n +7 \lt b^2$ and thus by taking square roots $x_{n+1} \lt b$

For a proof of ii)

We have that

$\displaystyle (x_{n+1})^2 - (x_n)^2 = -(x^2_n - x_n -7) = -(x_n-a)(x_n-b) \gt 0$ if $x_n \lt b$.

Thus $\displaystyle \{x_{n}\}$ is monotonically increasing and bounded above and so is convergent.

By setting $L = \sqrt{7+L}$, we can easily see that the limit is $\displaystyle b = \dfrac{1 + \sqrt{29}}{2}$

In fact, we can show that the convergence is linear.

$\displaystyle \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{b^2 - (7+x_n)}{(b+\sqrt{7+x_n})(b-x_n)} = \dfrac{1}{b + x_{n+1}}$

Thus $\displaystyle \lim_{n\to \infty} \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{1}{2b}$.

We can also show something a bit stronger:

Let $\displaystyle t_n = b - x_n$.

The we have shown above that $\displaystyle t_n \gt 0$ and $\displaystyle t_n \lt b^2$

We have that

$\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}$

Dividing by $\displaystyle b$ throughout we get

$\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}$

Using $\displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1$ we have that

$\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}$

And so

$\displaystyle \dfrac{t_n}{2b} \leq t_{n+1} \leq \dfrac{t_n}{b}$

This gives us that $\displaystyle b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$

  • $\begingroup$ This is years old... But I was wondering how you got the very last deduction? $\endgroup$
    – smalldog
    Apr 11, 2021 at 13:27
  • $\begingroup$ @acephalous: Sorry not sure what you are asking. And yeah it is pretty old so I might not be able to answer :) $\endgroup$
    – Aryabhata
    Apr 16, 2021 at 22:26
  • $\begingroup$ No problem if you can't... But I was wondering how you went from $\dfrac{t_n}{2b} \leq t_{n+1} \leq \dfrac{t_n}{b}$ to $b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$ (last sentence). $\endgroup$
    – smalldog
    Apr 17, 2021 at 7:54
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    $\begingroup$ Sort of telescoping: $t_{n+1} \le \frac{t_{n}}{b} \le \frac{t_{n-1}}{b^2} \le \frac{t_{n-2}}{b^3} \dots \le \frac{t_0}{b^{n+1}}$. Do the same for the left part and then use $t_n = b - x_n$ to get the inequalities for $x_n$. There might be off by one/two errors though. $\endgroup$
    – Aryabhata
    Apr 19, 2021 at 0:41

Hint: It morally converges to $\sqrt{7+\sqrt{7+\sqrt{7+.....}}}$. Call this limit $l$. Then, since the nested series of square roots extends indefinitely, $l=\sqrt{7+l}$

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    $\begingroup$ Superb. That's a beautiful argument. Thanks friend. $\endgroup$
    – user4082
    Nov 26, 2010 at 8:28
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    $\begingroup$ For this argument to work you need first to show that the sequence actually converges. $\endgroup$ Nov 26, 2010 at 15:18
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    $\begingroup$ Agreed; this answer should not have been accepted. $\endgroup$ Nov 26, 2010 at 16:24
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    $\begingroup$ @Qiaochu: It wasn't meant to be. I just intended it and marked it as hint. The OP just asked about the limit. $\endgroup$ Nov 26, 2010 at 16:26

HINT: Suppose the sequence converges to $x$. Then solve for $x = \sqrt{7+x}$


HINT: I just attended the case $x=6$, and by using similar tackling ways, you may easily solve the case $x=7$. Here you may find a possible approaching style, and pretty fast. Of course, this problem may be approached in various ways.


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