Simple recurrences converging to $\log 2, \pi, e, \sqrt{2}$ and so on See my question at the bottom of this post. The recurrence $P(n) x_{n+2} = Q(n)x_{n+1} - R(n)x_n$, where $P(n), Q(n), R(n)$ are  polynomials of degree $1$, sometimes leads to interesting results. Probably the most basic cases are:
For $\log\alpha$:
$$P(n) = \alpha (n+2), Q(n) = (2\alpha-1)(n+1)+\alpha, R(n)=(\alpha-1)(n+1)$$ $$\mbox{with } x_1=\frac{\alpha-1}{\alpha}, x_2 = \frac{(\alpha-1) (3\alpha-1)}{2\alpha^2}$$ 
We have $\lim_{n\rightarrow\infty} x_n = \log\alpha$. The convergence is fastest when $\alpha$ is close to $1$. The related recurrence
$$P(n) = 1, Q(n) = (2\alpha-1)(n+1)+\alpha, R(n)=(\alpha-1)\alpha(n+1)^2$$
$$\mbox{with } x_1=\alpha-1, x_2=(\alpha-1)(3\alpha-1)$$
yields 
$$\lim_{n\rightarrow\infty} \frac{x_n}{\alpha^n n!} = \log\alpha$$ and in addition $x_n$ is an integer if $\alpha>0$ is an integer.
For $\exp \alpha$:
$$P(n) = n+2, Q(n) = n+2+\alpha, R(n)=\alpha$$ $$\mbox{with } x_0=1, x_1 = 1+\alpha$$ 
We have $\lim_{n\rightarrow\infty} x_n = \exp\alpha$. The related recurrence
$$P(n) = 1, Q(n) = n+2+\alpha, R(n)=\alpha(n+1)$$
$$\mbox{with } x_0=1, x_1=1+\alpha$$
yields 
$$\lim_{n\rightarrow\infty} \frac{x_n}{n!} = \exp\alpha$$ and in addition $x_n$ is an integer if $\alpha$ is an integer.
For $\sqrt{2}$:
$$P(n) = 4(n+2), Q(n) = 6n+11, R(n)=2n+3$$ $$\mbox{with } x_0=1, x_1 = \frac{5}{4}$$ 
We have $\lim_{n\rightarrow\infty} x_n = \sqrt{2}$. The related recurrence
$$P(n) = n+2, Q(n) = 2(6n+11), R(n)=16(2n+3)$$
$$\mbox{with } x_0=1, x_1=10$$
yields 
$$\lim_{n\rightarrow\infty} \frac{x_n}{8^n} = \sqrt{2}$$ and in addition $x_n$ is an integer.
Comment
These formulas (and tons of other similar formulas) are easy to obtain, yet I could not find any reference in the literature. It would be interesting to see if one is available for $\gamma$ (the Euler Mascheroni constant), but I don't think so. Also, what happens when you change the initial conditions? What if you replace the recurrence by its equivalent differential equation, for instance 
$$(x+2) f(x) - (x+2+\alpha) f'(x) + \alpha f''(x) =0$$
corresponding to the case $\exp\alpha$?
Generalization to arbitrary initial values
As an example, here is what happens to the very first formula (the $\log \alpha$ case), if we change the initial conditions $x_1=\frac{\alpha-1}{\alpha}, x_2 = \frac{(\alpha-1) (3\alpha-1)}{2\alpha^2}$ to arbitrary values $x_1 = A, x_2=B$, assuming here that $\alpha=2$:
$$\lim_{n\rightarrow\infty} x_n = (5-8\log \alpha)\cdot A + (8\log \alpha -4) \cdot B.$$ 
You may try proving this formula. It was obtained empirically, I haven't proved it. And it works only if $\alpha = 2$.
For $\alpha \neq 2$, and also for the case $\sqrt{2}$, a general formula is 
$$\lim_{n\rightarrow\infty} x_n = c_1 A + c_2 B$$
where $c_1, c_2$ are constants not depending on the initial conditions. This might be a general property of these converging linear recurrences (at least those involving polynomials of degree one). Another property, shared by the converging systems described here, is as follows: 
$$A = B \Rightarrow \lim_{n\rightarrow\infty} x_n = A.$$
This implies that $c_1+c_2 = 1$.
How to obtain these recursions?
The case $\sqrt{2}$ can be derived from this other question. To me, it is the most interesting case as it allows you to study the digits of $\sqrt{2}$ in base 2. Some of these recursions can be computed with WolframAlpha, see here for the exponential case, and here for $\sqrt{2}$. Numerous other recurrences, with much faster convergence, can be derived from combinatorial sums featured in this WA article.
My question
I am looking for some literature on these linear, non-homogeneous second order recurrences involving polynomials of degree $1$. Also, I will accept any answer for a recurrence that yields $\pi$. Should be easy, using formulas (37) or (38) in this article as a starting point. 
If you find my question too easy, here is one that could be much less easy: change the initial conditions to $x_0=A, x_1=B$ in any of these formulas, and see if you can get convergence to a known mathematical constant.
 A: The generalized binomial theorem leads to rational powers of rationals.
$$(1+x)^p=1+px+\frac{p(p-1)x^2}{2}+\frac{p(p-1)(p-2)x^3}{3!}+\cdots$$
The recurrence relation between the terms is obvious.
Now, with $p=-1$, you get $\log(1+x)$ by term-wise integration, thus the logarithms of rationals. And substituting $x^2$ for $x$ and integrating, you obtain $\arctan(x)$, and $\pi$.
Finally, $e$ can be drawn by expanding
$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}+\cdots$$ and letting $n$ go to infinity. Here again, the recurrence is easy.
These series can also be seen as Taylor expansions of some functions, and the recurrence relations are those that link the derivatives evaluated at $0$. Hence you can apply this trick to functions defined by a differential equation.
E.g., let $y''=-y$, with $y(0)=1$ and $y'(0)=0$.
By induction, the even derivatives are $\pm1$ alternating and the odd ones are $0$. The terms of the Taylor expansion are
$$(-1)^n\frac{x^{2n}}{(2n)!},$$ which are such that $$t_{n+1}=-\frac{x^2}{(2n+1)(2n+2)} t_n$$ and with $x=1$, you get $\cos(1)$.
