Rewriting sequence from $X_{n+1}$ to $X_n$ I have the sequence:
$$
\begin{align}
X_{n+1} &= \frac{X_n^2 + 5}{2X_n} \\
X_1 &= 1
\end{align}
$$
I have to prove that it converges and find its limit after I write it in terms of $X_n$, which I can do, but I can't seem to convert $X_{n+1}$ in terms of $X_n$.
 A: As a first step, it may be worthwhile to calculate $X_2$, $X_3$, and perhaps a few other terms. For example, we get $X_2=3$ and $X_3=\frac{7}{3}$. After the jump to $3$, the terms appear to be decreasing. 
The sequence $(X_n)$ is obviously bounded below by $0$. If we can prove that from $n=2$ on it is decreasing, we will know that it converges. 
So we would like to prove that $X_n-X_{n+1}$ is positive when $n\ge 2$. Note that 
$$X_n-X_{n+1}=X_n -\frac{X_n^2+5}{2X_n}=\frac{X_n^2-5}{2X_n}.\tag{$1$}$$
If we can show that $X_n^2-5\gt 0$ for $n \ge 2$, then we will know from $(1)$ that the sequence is decreasing from $n=2$ on.
Consider the function  $f(t)=\frac{t^2+5}{2t}$ for positive $t$. By standard calculus, we can show that the minimum value of $f(t)$, for positive $t$, is $\sqrt{5}$. This value is taken on at $t=\sqrt{5}$. It is easy to verify that $X_n$ is never exactly $\sqrt{5}$. This shows that $X_n-X_{n+1} \gt 0$ if $n\ge 2$, and convergence follows. 
Remark: As has been pointed out by Ishan Banerjee, the recurrence is a very special one. It can be rewritten as 
$$X_{n+1}=\frac{1}{2}\left(X_n +\frac{5}{X_n}\right).$$
We may recognize this as the recurrence we get when we apply the Newton-Raphson procedure to the equation $x^2-5=0$. For square roots, the method was known in late Babylonian times. It is sometimes called the Heron Method.
The sequence $(X_n)$, as has been pointed out by Peter Tamaroff, converges to $\sqrt{5}$. We can exploit that knowledge to get sharp information about the behaviour of $(X_n)$. Note that
$$X_{n+1}-\sqrt{5}=\frac{X_n^2+5}{2X_n}-\sqrt{5}=\frac{X_n^2-2\sqrt{5}X_n+5}{2X_n}=\frac{(X_n-\sqrt{5})^2}{2X_n}.$$
For $n\gt 1$. We have $2X_n \gt 2\sqrt{5}$, and therefore 
$$0 \lt X_{n+1}\lt \frac{(X_n-\sqrt{5})^2}{2\sqrt{5}}$$
for all $n \gt 1$. 
What this means is that the sequence $(X_n)$ converges to $\sqrt{5}$ with extreme rapidity. If at a certain stage the error is, say, $\lt 10^{-4}$, then at the next stage the error is $\lt \frac{10^{-8}}{2\sqrt{5}}$. 
A: We calculate $X_2=3$, $X_3=\frac{7}{3}$ and $X_4=\frac{47}{21}<X_3$.
Then we prove by induction that $\sqrt{5}\leq X_n\leq \frac{7}{3}$ for all $ n\geq 3$.
If we denote $f$ the function $x\mapsto \frac{x^2+5}{2x}$, we have $X_{n+1}=f(X_n)$, and since $f'(x)=\frac{2(x^2-5)}{4x^2}$, $f$ is increasing for all $x\geq \sqrt{5}$.
Hence, since $X_3\geq X_4\geq \sqrt{5}$, we have $X_n\geq X_{n+1}, \forall n\geq 3$ by induction.
We conclude that the sequence $(X_n)_{n\geq 3}$ is decreasing and bounded below by $\sqrt{5}$, hence it's convergent to the positive solution of equation $f(x)=x$, i.e. to $\sqrt{5}$.
A: *

*$X_{n+1}= \frac{X_n}{2} + \frac{5}{2X_n} \geq 2 \sqrt{\frac{X_n}{2} \cdot \frac{5}{2X_n}} = \sqrt{5}$

*It could be proven that if $X_n \geq \sqrt 5$, then $X_{n+1} - X_n = \frac{5-X_n^2}{2X_n}\leq 0$, which means $X_n$ is decreasing sequence, with a lower bound of $\sqrt{5}$. Therefore $\lim X_n$ exists, note it as $X_0$

*$X_0 = \lim X_{n+1} = \lim\frac{X_n^2 +5}{2X_n} = \frac{X_0^2 +5}{2X_0}$, solving this equation gets $X_0 = \sqrt 5$. ($X_0 = -\sqrt 5$ is obviously not the limit cuz $X_n >=0$)

