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This sequence is certainly limited \begin{equation} a_n=\frac{2+\sin(n+1)}{2+\sin(n)} \end{equation} For example, if $\sin(n+1)=1$ and $\sin(n)=-1$, we have: $a_n\leq 3$. But I believe that there is $0<M<3$ such that $a_n\leq M$.

I would like to find this value of M. Who has any suggestions for me?

Thank you very much.

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up vote 3 down vote accepted

Fix some $\alpha$ and, for every real number $x$, let $$a(x)=\dfrac{2+\sin(x+\alpha)}{2+\sin(x)}. $$ Using the addition formula $\sin(x+\alpha)=\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x)$, one sees that $a(x)\leqslant M$ if and only if $$ \sin(\alpha)\cos(x)+(\cos(\alpha)-M)\sin(x)\leqslant2M-2. $$ There exists $(r,t)$, $r\gt0$, such that $\sin(\alpha)=r\sin(t)$ and $\cos(\alpha)-M=r\cos(t)$, thus the LHS is $r\sin(x+t)\leqslant r$. This shows that if $2M-2\geqslant r$ then $a(x)\leqslant M$ for every real number $x$.

Assume that $M\geqslant1$. Then $r^2=\sin(\alpha)^2+(\cos(\alpha)-M)^2$ hence the condition is $$ 4(M-1)^2\geqslant1-2\cos(\alpha)M+M^2,\tag{$\ast$} $$ that is, $$ 3M^2-2(4-\cos(\alpha))M+3\geqslant0,\qquad M\geqslant1. $$ Numerically, this means one can choose $$ M_\alpha=\frac13\left(4-\cos(\alpha)+\sqrt{7-8\cos(\alpha)+\cos(\alpha)^2}\right). $$ Thus, $1\leqslant M_\alpha\leqslant3$ for every $\alpha$ and $M_\alpha\lt3$ for every $\alpha$ not in $\pi+2\pi\mathbb Z$. In particular, $M_1\lt1.73$.

To show more modestly that $M=2$ works when $\alpha=1$, it suffices to check that, for $M=2$, $(\ast)$ reads $\cos(1)\geqslant\frac14$, and to note that this last inequality is obvious since $1\leqslant\frac\pi3$ and $\cos(\frac\pi3)=\frac12$ hence $\cos(1)\geqslant\frac12$.

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