On convergence of a sequence $\{x_n\}$ , given that the sequence $\{x_{n+1}+f(x_n)\}$ converges for some function $f$ on real line If $\{x_n\}$ is a real sequence such that the sequence $\{2x_{n+1}+\sin x_n\}$ is convergent , then is it true that $\{x_n\}$ convergent ? 
 A: Say $b = \lim 2x_{n+1} + \sin x_n$ and $a$ is the (unique) real number such that $2a + \sin a = b$. Then
$$0 = \lim_{n\to\infty} (2x_{n+1} + \sin x_n - 2a - \sin a) = \lim_{n\to\infty} \biggl(2(x_{n+1}-a) + 2\cos\frac{a+x_n}{2}\sin \frac{x_n-a}{2}\biggr).$$
Define $u_n = x_n - a$, then
$$v_n := u_{n+1} + \cos(a + u_n/2)\sin (u_n/2) \to 0.$$
We would like to conclude that $u_n \to 0$. Suppose that weren't the case, and
$$\eta := \limsup_{n\to \infty}\: \lvert u_n\rvert > 0.$$
We can without loss of generality assume that $\lvert v_n\rvert < \eta/8$ for all $n$. For every $n$ such that $\lvert u_{n+1}\rvert > 7\eta/8$, it then follows that
$$\frac{3}{4}\eta < \lvert u_{n+1} - v_n\rvert = \lvert \cos(a+u_n/2)\sin(u_n/2)\rvert \leqslant \frac{\lvert u_n\rvert}{2},$$
so
$$\lvert u_n\rvert > \frac{3}{2}\eta.$$
Since there are arbitrarily large $n$ such that $\lvert u_{n+1}\rvert > 7\eta/8$, we obtain the contradiction
$$\eta = \limsup_{n\to\infty}\: \lvert u_n\rvert \geqslant \frac{3}{2}\eta.$$
A: By hypothesis, the sequence $(y_n)$ defined by $y_n=2x_{n+1}+\sin x_n$ converges to some finite limit. Furthermore, $x_{n+1}=\frac12y_n-\frac12\sin x_n$ and $-1\leqslant\sin x_n\leqslant1$ for every $n$, hence there exists $\ell_0\leqslant1.01$ and $a_0$ such that, for every $n$ large enough, $a_0\leqslant x_n\leqslant a_0+\ell_0$.
Now, assume that, for every $n$ large enough, $a_k\leqslant x_n\leqslant a_k+\ell_k$ for some given $\ell_k$ and $a_k$. Then, the image of any interval of length $\ell_k$ by the sine function has length at most $\ell_k$ hence, using the identity $x_{n+1}=\frac12y_n-\frac12\sin x_n$ once again, one sees that  there exists some $\ell_{k+1}\leqslant0.51\ell_k$ and $a_{k+1}$ such that, for every $n$ large enough, $a_{k+1}\leqslant x_n\leqslant a_{k+1}+\ell_{k+1}$.
The intersection $\bigcap\limits_k[a_k,a_k+\ell_k]$ is reduced to at most one point since $\ell_k\to0$. It remains to note that $(x_n)$ is bounded, hence $(x_n)$ has at least one limit point, to deduce that $(x_n)$ converges.

This approach proves the following more general result, whose statement might help to pinpoint the features of the present one that make it hold:

Let $y_n=cx_{n+1}+f(x_n)$, for some fixed $c$ and $k$-Lipschitz function $f$, with $k<|c|$. If the sequence $(y_n)$ converges, then the sequence $(x_n)$ converges.

