# The Sum of $\sum\limits_{n=1}^\infty \left(\sin\frac{10}{n} -\sin\frac{10}{n+1}\right)$

$$\sum\limits_{n=1}^\infty \left(\sin\frac{10}{n} - \sin\frac{10}{n+1}\right)$$

I see that as $n$ approaches $\infty$ that it'll be 0 thus convergent. However, I'm unclear of the manipulation that's implemented to get an actual result ($\sin(10) = -0.5442$). Can these $\sin$ elements be put together in such a way to be able to evaluate $\sin$ at some real number or is there a need for something else to be done?

• The fact that $a_n$ has a limit of $0$ as $n\to\infty$ doesn't necessarily mean $\sum a_n$ is convergent. Consider $\sum_n \frac{1}{n}$ – apnorton Feb 21 '15 at 23:02

Just telescope it: look at the few initial terms: $$\sin 10 - \sin 5 + \sin 5 - \sin(10/3) + \sin(10/3)- \sin(10/4)+\cdots.$$ Meaning: $$\sum_{k=1}^n \left(\sin \frac{10}{k}-\sin\frac{10}{k+1}\right) = \sin 10 - \sin\frac{10}{n+1}.$$This way: \begin{align}\sum_{n=1}^{+\infty}\left(\sin\frac{10}{n} - \sin\frac{10}{n+1}\right) &= \lim_{n \to +\infty} \sum_{k=1}^n \left(\sin\frac{10}{k} - \sin\frac{10}{k+1}\right)\\ &= \lim_{n \to +\infty}\left(\sin 10 - \sin\frac{10}{n+1}\right) = \sin 10. \end{align}
• Try to remember it anytime you see something like $\sum (f(n) - f(n+1))$ or $\sum (f(n) - f(n-1))$, it's a neat trick. – Ivo Terek Feb 21 '15 at 23:06