Suppose that $x_0 \in [0,\infty)$ and $x_{n+1} =\sqrt{\frac{3x_n+2}{2}}$. Demonstrate that $ (x_n)_{n \in \mathbb {N}}$ converges and compute $\lim_\limits{n\to\infty} x_n$.

We've not been taught yet how to use L'Hopital or Taylor expansion or asymplotes; I'm in third year of highschool and we've been taught about limits for half a year now, don't expect a hard way to solve this. It's a problem from my test.

  • $\begingroup$ In what region of $[0, \infty)$ does $x < \sqrt{\frac{3x+2}{2}}$ hold, and in what region does $x > \sqrt{\frac{3x+2}{2}}$ hold? $\endgroup$
    – Michael L.
    Commented Jan 17, 2018 at 14:37
  • 3
    $\begingroup$ I usually like to do it in reverse order. That is, first, assuming that the sequence does converge, find what it could possibly converge to. Then, once I know that, it's often easier to show that the sequence indeed converges. $\endgroup$
    – Arthur
    Commented Jan 17, 2018 at 14:38

2 Answers 2


Hint: The sequence converges to $2$. To guess the limit take $n\rightarrow \infty$ at both sides of the recursion, and assume $x_n$ is converging. Since, $x_n$ converges the LHS is equal to $a$, and the RHS is equal to $\sqrt{\dfrac{3a+2}{2}}$. Solve this equation to find the candidate limits.

Now, for the convergence.

If $x_0\in [0,2)$, show by induction that $x_n<2$ for all $n$. Once you have establish that, show, by induction, that $x_n$ is increasing, hence $x_n$ must be a convergent sequence.

The case $x_0\in(2,\infty)$ can be handed similarly.

Last case $x_0=2$ is the easiest.



  • $\displaystyle x_{n+1} - x_n = \frac{\sqrt{3x_n+2} - \sqrt{3x_{n-1}+2}}{\sqrt{2}} = \frac{3(x_n-x_{n-1})}{\sqrt{2}\left(\sqrt{3x_n+2} + \sqrt{3x_{n-1}+2}\right)}\,$, therefore the differences between consecutive terms have the same sign, so the sequence is monotonic;

  • $\displaystyle x_{n+1} =\sqrt{\frac{3x_n+2}{2}} \iff 2x_{n+1}^2=3x_n+2=0 \iff 2(x_{n+1}^2-4) = 3(x_n-2)\,$, therefore $x_{n+1}-2$ and $x_n-2$ have the same sign, so all terms are on the same side of $2$;

  • $\displaystyle x_1 >= x_0 \iff 2x_0^2-3x_0-2 = (x_0-2)(2x_0+1) \le 0 \iff x_0 \le 2\,$, therefore:

    • for $x_0 \le 2$ the sequence is increasing and bounded above by $2$;

    • for $x_0 \gt 2$ the sequence is decreasing and bounded below by $2$.

[ EDIT ]   For the sake of the belated downvoter, here is the last hint:

  • the sequence converges since it is bounded and monotonic, then let $\,x\,$ be its limit; passing the recurrence relation to the limit gives $\,x = \sqrt{\frac{3x+2}{2}} \implies 2x^2-3x-2=0\,$ with $\,x \ge 0\,$.
  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$
    – dxiv
    Commented Feb 26, 2018 at 19:10

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