<s>[INST] Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
 [/INST] Using algebra
Rewriting
$$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)$$ as
$$a_{n+1}=a_n-\left(a_n-\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)\right)=a_n-\frac{a_n^2-3}{2 a_n}$$ we can recognize the itegrative scheme for solving $a^2=3$; so the limit is $\sqrt 3$.
Considering the function $f(a)=a^2-3$, starting with $a_0=1$, we have
$f(1)=-2$ and $f''(1)=2$, $f(1)\times f''(1)=-4<0$, then, by Darboux theorem, we shall have one (and only one overshoot of the solution). This corresponds to $a_2$ and, later, we shall converge quadratically to the root.
$$\left(
\begin{array}{cc}
n & a_n \\
 1 & 1.0000000000000000000 \\
 2 & 2.0000000000000000000 \\
 3 & 1.7500000000000000000 \\
 4 & 1.7321428571428571429 \\
 5 & 1.7320508100147275405 \\
 6 & 1.7320508075688772953 \\
 7 & 1.7320508075688772935
\end{array}
\right)$$
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