1
$\begingroup$

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

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?

$\endgroup$
  • 2
    $\begingroup$ The fact that $a_n$ has a limit of $0$ as $n\to\infty$ doesn't necessarily mean $\sum a_n$ is convergent. Consider $\sum_n \frac{1}{n}$ $\endgroup$ – apnorton Feb 21 '15 at 23:02
7
$\begingroup$

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: $$\begin{align}\sum_{n=1}^{+\infty}\left(\sin\frac{10}{n} - \sin\frac{10}{n+1}\right) &= \lim_{n \to +\infty} \sum_{k=1}^n \left(\sin\frac{10}{k} - \sin\frac{10}{k+1}\right)\\ &= \lim_{n \to +\infty}\left(\sin 10 - \sin\frac{10}{n+1}\right) = \sin 10. \end{align}$$

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Ivo Terek Feb 21 '15 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.