How to solve $x_{100}$ with $x_{n+1}=\frac{2x_n}{2+x_n}+1$ and $x_{1}=3$ How to solve $x_{100}$ with $x_{n+1}=\frac{2x_n}{2+x_n}+1$ and $x_{1}=3$?

Can anybody shed light on this? regards.
tl;dr: considering a related sequence will help you obtain (very easily) the closed-form for $x_n$, from which you can plug in $n=100$ and get
$x_{100} = 2 + \frac{3}{2^{200}-1} \simeq 2 + 2\cdot 10^{-60}.$



Details: It is pretty clear that the sequence converges to $2$ (if you do not know how to show it, prove by induction it is monotone decreasing and positive, hence convergent; and show the only possible limit is then $\ell=2$).

So let us consider instead the positive sequence $(a_n)_{n\geq 1}$ defined by $a_n \stackrel{\rm def}{=} x_n -2$ for all $n\geq 1$, which now converges to $0$ and satisfies the recurrence relation (from the one that $(x_n)_n$ satisfies)
$$
a_{n+1} = \frac{a_n}{a_n+4} = \frac{1}{1+\frac{4}{a_n}}, \qquad n\geq 1 \tag{1}
$$
with $a_1 = 1$.

Mmh, there are a lot of reciprocals there, and they are not that nice. Let us go further defining the (new) positive sequence $(b_n)_{n\geq 1}$ by
$b_n \stackrel{\rm def}{=} \frac{1}{a_n} = \frac{1}{x_n -2}$, so that
$$
b_{n+1} = 1+{4}{b_n}, \qquad n\geq 1 \tag{2}
$$
with $b_1 = 1$.

This looks much nicer. Solving this should not be too hard: there are various standard methods, and they will lead to
$$
b_n = \frac{4^n-1}{3}, \qquad n\geq 1
$$
from which you eventually get the closed-form expression for $x_n$:
$$
x_n = \frac{1}{b_n}+2 = \frac{3}{4^n-1}+2, \qquad n\geq 1. \tag{3}
$$
(Sanity check: it does converge to $2$, is positive, and decreasing with $x_1=3$.)

And finally: from (3), you get
$$
x_{100} = 2 + \frac{3}{2^{200}-1}.
$$
you are trying to find the fixed point of the function f(x)=(2x/(1+x))+1 with a derivative 2/(1+x)^2&lt;1 on the domain [1,10],say. The fixed point is 2. you can find the term you are after by consulting any book on numerical analysis ((Mendelson's book)ang go from  there to the fixed point algorithm.