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Let $$\lim\limits_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]}$$ exists.

What are the value of the limitation and $a$

My attempt $$\lim\limits_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]}=0$$

Let $l=\lim\limits_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]}$, if $l\neq0$,then we have, $$\sin((n+1)a)-\sin(na)\leq|\sin((n+1)a)|+|\sin(na)|\leq2$$ $l\leq2$, but

$$\lim_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]}=\lim_{n\to\infty}{\left[\sin(na)-\sin((n-1)a)\right]}$$

so $$\lim_{n\to\infty}{\left[\sin((n+1)a)-\sin((n-1)a)\right]}=2l$$ we get, $$\lim_{n\to\infty}{\left[\sin((n+1)a)-\sin((n-k)a)\right]}=kl$$ $\exists k\in\mathbb{N},kl>2$, so $l=0$.

But I don't know the value of $a$, so I want to get some help.

Thanks

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    $\begingroup$ Notice how you said "if $l \neq 0$ then we have". Then you conclude $l=0$ $\endgroup$ – Hugh Feb 23 '19 at 12:59
  • $\begingroup$ Use the difference of sines formula to write $\sin A - \sin B$ as a product of a sine and a cosine of two other quantities. $\endgroup$ – Teresa Lisbon Feb 23 '19 at 13:00
  • $\begingroup$ @Hugh What do you mean? $\endgroup$ – Li Taiji Feb 23 '19 at 13:23
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    $\begingroup$ The argument works if you assume $l>0$ and you should modify it for the case $l<0$ (not difficult). $\endgroup$ – egreg Feb 23 '19 at 14:53
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\begin{align} L &= \lim\limits_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]} = \lim\limits_{n\to\infty}{\left[2\cos\left(na+\frac a2\right)\sin\left(\frac a2\right)\right]} \\ & = 2\sin \left(\frac a2\right)\lim\limits_{n\to\infty}{\cos\left(\frac a2+na\right)}. \end{align} Hence, $a=k\pi, k\in \mathbb Z$ and $L=0$.

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  • $\begingroup$ With $a = 2 \pi$ and $n = 1$ we have $\cos\left( \frac{a}{2} + n a \right) = \cos(3 \pi) \neq 0$, so why is the limit zero? $\endgroup$ – Ramanujan Feb 23 '19 at 23:10
  • $\begingroup$ @Viktor Glombik, because for $a=2\pi$, the sine is zero. $\endgroup$ – farruhota Feb 23 '19 at 23:48
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Hint: Use that $$\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$

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  • $\begingroup$ $a=k\pi,k\in\mathbb{Z}$? $\endgroup$ – Li Taiji Feb 23 '19 at 13:44

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