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?


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 $k=1,2,3,\cdots,$ $$ \sin(\pi k)=0, \qquad \cos(\pi k)=(-1)^k. $$

  • $\begingroup$ 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

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