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It is known that the sequence $\sin(1), \sin(2),\sin(3), \cdots$ is divergent.

In book Yet Another Introduction to Analysis by V. Bryant (p.240), the author proves in following lines:

(1) There exists increasing sequence $k_1,k_2,\ldots$ of integers such that $\sin(k_i)\ge \frac{1}{2}$.

(2) There exists increasing sequence $j_1,j_2,\ldots$ of integers such that $\sin(j_i)\le -\frac{1}{2}$.

Using these facts, it is easy to arrive divergence of $\sin(1),\sin(2),\sin(3),\ldots$.

Q. Suppose $a$ is a given real number in $(-1,1)$. Is it true that there exists an infinite sequence $c_1,c_2,c_3,\ldots$ of integers such that $\sin(c_i)\ge a$? (This question comes from the numbers $1/2$ or $-1/2$ used in the arguments by Bryant.)

Beginner
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1 Answers1

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The answer is "yes", any other $a$ in $(-1, 1)$ will work. However, $a = \frac12$ (and $-\frac12$ for case 2) is a strategic choice, because it's a lot easier to prove than for $a$ closer to $1$.

The first interval (for positive $x$) where $\sin(x)\geq\frac12$ is $\left[\frac\pi6, \frac{5\pi}6\right]$. The width of this interval is $\frac{2\pi}3>1$, so we are guaranteed to find an integer in there. Let $k_1$ be an integer in that interval. The next interval also has width greater than $1$, so we are guaranteed to find an integer in there too. We can let $k_2$ be one of those. And so on.

For $a$ close enough to $1$, the width of the corresponding interval becomes lower than $1$, and for such choices of $a$ it is no longer the case that each corresponding interval like above contains an integer. In fact, it is not immediate that any such interval contains an integer, much less infinitely many of them.

So we need a new argument to prove that there is always another interval with an integer in it (even though it may not be the next one). Such a proof will hinge on the fact that $2\pi$ is irrational, since that's the distance between the start of one interval and the start of the next. This irrationality implies that there will always be another interval that starts close enough to an integer to include that integer.

Arthur
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