Iteration on sequences I am trying to find a general formula for this sequence, in order to find its limit (has to be this way):
$$X_{k+1} = \frac {X_k} 2 + \frac 1 {X_k}$$
I cannot get a grasp on it; after $2$ or $3$ iterations, the sequence gets huge and I don't see the pattern.
Any help is greatly appreciated!
 A: If $X_0=\sqrt 2$, then the sequence is constant, so we can assume $X_0\not = \sqrt 2$. We may also assume that $X_0>0$, as the negative case is the same.
Let $X_0=\sqrt 2\cdot\frac {r+1}{r-1} $, or $r=\frac{X_0+\sqrt2}{X_0-\sqrt2}$, then the closed form is $X_k=\sqrt 2\cdot\frac {r^{2^k}+1}{r^{2^k}-1}$. The way to get the closed form is as follows.
Dividing by $\sqrt 2$:
$$\frac {X_{k+1}}{\sqrt 2}=\frac 12(\frac {X_{k}}{\sqrt 2}+\frac {\sqrt 2}{X_{k}})$$
Substituting $\frac {X_{k}}{\sqrt 2}=S_k+1$:
$$S_{k+1}+1=\frac 12 \cdot(S_k+1+\frac 1{S_k+1})$$
$$S_{k+1}=\frac 12 \cdot(S_k-1+\frac 1{S_k+1})$$
$$S_{k+1}=\frac 12 \cdot\frac {S_k^2}{S_k+1}$$
Taking the inverse:
$$\frac 1{S_{k+1}}=2  \cdot\frac {S_{k}+1}{S_k^2}$$
$$\frac 1{S_{k+1}}= 2\cdot (\frac 1{S_{k}})^2+2\cdot \frac 1{S_{k}}$$
Completing the square:
$$\frac 1{S_{k+1}}+\frac 12=2\cdot(\frac 1{S_{k}}+\frac 12)^2$$
$$\frac 2{S_{k+1}}+1=(\frac 2{S_{k}}+1)^2$$
This means each term is the square of the previous term, which we can denote as  $\frac 2{S_{k}}+1=r^{2^{k}}$. Substituting everything back into $X_k$ then we have the close form.
Proving that the closed form is actually correct:
\begin{align}
\frac {X_k}2+\frac 1{X_k}
&=\frac 1{\sqrt2} \cdot (\frac {r^{2^k}+1}{r^{2^k}-1}+\frac {r^{2^k}-1}{r^{2^k}+1})\\
&=\frac 1{\sqrt2} \cdot  \frac {(r^{2^k}+1)^2+({r^{2^k}-1})^2}{(r^{2^k}-1)({r^{2^k}+1})} \\
&=\frac 1{\sqrt2}\cdot  \frac {r^{2^{k+1}}+2r^{2^k}+1+r^{2^{k+1}}-2r^{2^k}+1}{r^{2^{k+1}}-1}\\
&=\sqrt 2\cdot  \frac {r^{2^{k+1}}+1}{r^{2^{k+1}}-1}\\
&=X_{k+1}
\end{align}
A: If the sequence converges, it must converge to an $x$ such that
$$x=\frac x2+\frac1x$$ or $$x^2=2.$$
Then with $x_k:=y_k+\sqrt2$,
$$y_{k+1}+\sqrt 2=\frac{y_k+\sqrt2}2+\frac1{y_k+\sqrt2}$$ and
$$y_{k+1}=\frac{y_k^2}{2(y_k+\sqrt2)}\le\frac{y_k^2}{2\sqrt2}.$$
You easily see that starting from $y_0<1$, you get a (quickly) decreasing sequence with limit $0$.
