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 ? 

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
$$\eta = \limsup_{n\to\infty}\: \lvert u_n\rvert \geqslant \frac{3}{2}\eta.$$
