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Consider:

$$\lim\limits_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}=1$$

I figured out that $$\lim\limits_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}\leqslant\lim_{n\to +\infty}\sqrt[n]{2}=1$$

What about the other side inequality?

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4 Answers 4

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We show that $|\sin(n+1)|+|\sin(n)|$ is never tiny, by finding a crude lower bound.

Suppose that $|\sin(n)|\lt 0.3$. By the addition formula, we have $\sin(n+1)=\sin n \cos 1+\cos n\sin 1$. Thus $$|\sin(n+1)|\ge |\cos n\sin 1|-|\sin n\cos 1|.\tag{1}$$ We have $|\cos n|\gt \sqrt{1-(0.3)^2}\gt 0.8$ and since $\sin 1\gt 0.8$ we have $|\cos n\sin 1|\gt 0.64$. But $|\sin n\cos 1|\lt 0.3$, so from (1) we conclude that $|\sin(n+1)|\gt 0.3$.

We have shown that one at least of $|\sin(n+1)|$ and $|\sin(n)|$ is $\ge 0.3$, so the sum is greater than $0.3$.

Finally, $\sqrt[n]{0.3}$ has limit $1$ as $n\to\infty$, and the squeeze is on.

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You have $2>|\sin(n+1)|+|\sin(n)|>0$ is continuous and periodic, so there is an $a$ witch satisfies $a>0$ and $2>|\sin(n+1)|+|\sin(n)|>a$, now you can take the n-th root of a, bacause $\lim_{n\to +\infty}\sqrt[n]{a}=1$

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  • $\begingroup$ Great answer, just want to add that the 'satisfies $2> a > 0$ is superfluous. Your second inequality is stronger, and can be established using the fact that $|sin(n+1)| + |sin(n)|$ is continuous and periodic. $\endgroup$ Oct 25, 2015 at 9:34
  • $\begingroup$ @moorish yeah, once it's periodic and strictly positife (and well defined evrywhere), it is bounded with strictly positive nubers, so we have it immediatly. $\endgroup$
    – Baconaro
    Oct 25, 2015 at 9:37
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    $\begingroup$ By a more detailed analysis, $\sin(1)\le|\sin(n+1)|+|\sin(n)|\le 2\sin((\pi-1)/2)$, and this is tight. $\endgroup$
    – user65203
    Oct 25, 2015 at 9:40
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It is enough to study $f(x)=\left|\sin(x)\right|+\left|\sin(x+1)\right|$ over $[0,\pi]$ to be able to state that: $$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\geq \sin(1). $$ Equality is attained at $x\in\pi\mathbb{Z}$ and at $x\in -1+\pi\mathbb{Z}$.

We may also improve the upper bound in the following way: $$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\leq 2\cos\left(\frac{1}{2}\right). $$ Equality is attained at $x=\frac{\pi-1}{2}+\pi\mathbb{Z}$.

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  • $\begingroup$ There is actually no need to study $f(x)$ at all. It is enough to prove that it doesn't vanish on $[0,\pi]$, and then it automatically has a positive lower bound by compactness. $\endgroup$ Oct 25, 2015 at 22:04
  • $\begingroup$ @YuvalFilmus: good point. Do you wonder writing a solution based on that insight? $\endgroup$ Oct 25, 2015 at 23:17
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It is known that the zeroes of $\sin$ are at the point $n\pi$ for integer $n$. In particular, $\sin x$ and $\sin(x+1)$ cannot both be zero, that is, $|\sin x| + |\sin (x+1)| > 0$. The function $f(x) = |\sin x| + |\sin (x+1)|$ is positive and periodic with period $2\pi$, so due to compactness it attains its minimum somewhere in $[0,2\pi]$, let's say at the point $x_0$. Thus $f(x) \geq f(x_0) > 0$ for all $x$, and the limit follows.

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