2
$\begingroup$

I know that the following series

$$\sum_{k=1}^{\infty} \sin\left(\pi\big(k+\frac{1}{k}\big)\right)$$

is alternating and that it converges, but I have a question regarding the alternating series test.

The test basically says that for an alternating series, the series converges if $a_k \rightarrow 0$ as $k \rightarrow \infty$ and $a_k$ is monotonically decreasing.

Normally, an alternating series would be written in the form

$$\sum_{k=1}^{\infty} (-1)^k a_k $$

or something similar to that. But in the case with $\sum_{k=1}^{\infty} sin(\pi(k+\frac{1}{k})$, I simply don't know which part of it $a_k$ is. Any advice regarding this?

$\endgroup$
5
$\begingroup$

You may observe that $$ \sin\left(\pi\big(k+\frac{1}{k}\big)\right)=\sin(\pi k)\cos\left(\frac{\pi}{k}\right)+\sin\left(\frac{\pi}{k}\right)\cos(\pi k)=0+(-1)^k\sin(\frac{\pi}{k}) $$ using, for $k=1,2,3,\cdots,$ $$ \sin(\pi k)=0, \qquad \cos(\pi k)=(-1)^k. $$

$\endgroup$
  • $\begingroup$ That is very helpful, however, treating $sin(\frac{\pi}{k})$ as $a_k$, is $a_k$ then really monotonically decreasing? $\endgroup$ – m.bing Dec 3 '15 at 20:47
  • $\begingroup$ @m.bing Good question. The answer is yes. Setting $f(x)=\sin\left(\frac{\pi}x\right)$, you have $f'(x)=-\frac{\pi}{x^2}\cos\left(\frac{\pi}x\right)< 0$ for all $x > 2$ thus $f$ is decreasing over $(2,+\infty)$. $\endgroup$ – Olivier Oloa Dec 3 '15 at 21:00
  • $\begingroup$ Ah, thank you, now I think I get it. :) $\endgroup$ – m.bing Dec 3 '15 at 21:04
  • $\begingroup$ I'm sorry, I thought I had it but I lost it apparantly. Why can the first expression be written $\sin\left(\pi k\right) \cos\left(\frac{\pi}{k}\right) + \sin\left(\frac{\pi}{k}\right) \cos\left(\pi k\right) $? I thought I should use a trig identity but it didn't work for me after all... $\endgroup$ – m.bing Dec 3 '15 at 22:26
  • $\begingroup$ @m.bing I've used $\sin (a+b)=\sin a \cos b+\sin b \cos a$. Obtained, for example, by identifying $$\sin (a+b)= \Im \left(e^{i(a+b)}\right)= \Im \left(e^{ia}\times e^{ib}\right)=\Im \left((\cos a +i \sin a)(\cos b +i \sin b)\right)$$ $\endgroup$ – Olivier Oloa Dec 3 '15 at 22:29

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.