3

https://math.stackexchange.com/a/1119137/29156

Shouldn't Abel's answer be:

we will use the fact that $$2 \sin 1 \sin k = \cos(k-1) -\cos(k + 1)$$let $S = \sin 1 + \sin 2 \cdots + \sin n,$ then

$\begin{align} 2S \sin 1 &= 2\sin 1 \sin 1+ 2 \sin 1 \sin 2 + 2 \sin 1 \sin 3\cdots +2 \sin 1 \sin n \\ &=(1 - \cos 2) +(\color{red}{\cos 1} - \cos3) +(\color{red}{\cos 2} - \cos 4)+\cdots +(\cos (n- 1) - \cos ( n + 1) )\\ &=1 + \color{red}{\cos 1} - \color{red}{\cos (n)} - \color{red}{\cos(n+1)} \end{align}$

?

Obviously this is still bounded

Adam Rubinson
  • 20,052

1 Answers1

2

hint

Yes, Abel's result seems to be false. $$2S\sin(1)=\sum_{k=1}^n\Big((\cos(k-1)-\cos(k))+(\cos(k)-\cos(k+1))\Bigr)=$$

$$1-\cos(n)+\cos(1)-\cos(n+1)=$$

$$2\sin^2(\frac{n+1}{2})+\cos(1)-\cos(n)=$$

$$2\sin(\frac{n+1}{2})\Bigl(\sin(\frac{n+1}{2})+\sin(\frac{n-1}{2})\Bigr)=$$

$$4\sin(\frac{n+1}{2})\sin(\frac n2)\cos(\frac 12)$$

Thus

$$\boxed{S=\frac{\sin(\frac{n+1}{2})\sin(\frac n2)}{\sin(\frac 12)}}$$

  • If I plug in $n=3$ in that final formula I get approximately $0.07,$ which is surely not equal to $\sin 1 + \sin 2 + \sin 3.$ But I also get inconsistent results from Abel's formula. – David K May 13 '22 at 16:47
  • Your result is wrong also. Check when $n=0$ it should be $S=0$. – Somos May 13 '22 at 18:32
  • Good now. Easy to make simple mistakes unless you check. – Somos May 13 '22 at 22:10