# The value of $\lim\limits_{n\to\infty}{\left[\sin((n+1)a)-\sin(na)\right]}$

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

• Notice how you said "if $l \neq 0$ then we have". Then you conclude $l=0$ – Hugh Feb 23 '19 at 12:59
• 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. – Teresa Lisbon Feb 23 '19 at 13:00
• @Hugh What do you mean? – Li Taiji Feb 23 '19 at 13:23
• The argument works if you assume $l>0$ and you should modify it for the case $l<0$ (not difficult). – egreg Feb 23 '19 at 14:53

\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 $$L=0$$.
• 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? – Ramanujan Feb 23 '19 at 23:10
• @Viktor Glombik, because for $a=2\pi$, the sine is zero. – farruhota Feb 23 '19 at 23:48
Hint: Use that $$\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$
• $a=k\pi,k\in\mathbb{Z}$? – Li Taiji Feb 23 '19 at 13:44