Help please to prove the convergence: $$\sum_{n=1}^{\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)\cos \frac{\pi n(n-1)}2$$

It can be proved with Dirichlet's test, but there are come problems with $$\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)$$ monotone. Other steps in this way are clear. Or I should use another test?

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    $\begingroup$ Do you know Taylor series? $\endgroup$ – Clayton Oct 16 '17 at 16:59
  • $\begingroup$ Yes, what you want to decompose? $\endgroup$ – Yury Gladkikh Oct 16 '17 at 17:05
  • $\begingroup$ Third installment of the same question. Did you warn the potential answerers that their answers will disappear when you will delete this and switch to the fourth installment? $\endgroup$ – Did Oct 16 '17 at 17:51
  • $\begingroup$ This is the last question, sorry :) Older installments were a bit different and they didn't answer the question. I hope it will. $\endgroup$ – Yury Gladkikh Oct 16 '17 at 17:56
  • $\begingroup$ "Older installments were a bit different and they didn't answer the question" Seriously? How were they different? And why would you avoid deleting the answers you receive this time since you already did exactly this, twice? $\endgroup$ – Did Oct 16 '17 at 18:22

Let $\frac{1}{n}=x$.

Hence, $$\lim_{n\rightarrow\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)=\lim_{x\rightarrow0}\left(\frac{1}{\sin{x}}-\frac{\cos{x}}{x}\right)=$$

$$\lim_{x\rightarrow0}\frac{x-\frac{1}{2}\sin2x}{x\sin{x}}=\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin{x}+x\cos{x}}=$$ $$=\lim_{x\rightarrow0}\frac{2\sin^2x}{x\left(\frac{\sin{x}}{x}+\cos{x}\right)}=0.$$

Since with $$\sum_{n=1}^{\infty}\cos\frac{\pi n(n-1)}{2}$$ you know what to do, by the Dirichlet's Test we'll done if we'll prove that $a_{n+1}\leq a_{n}$, where $$a_n=\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n$$ or $\lim\limits_{x\rightarrow0^+}f'(x)>0$, where $$f(x)=\frac{1}{\sin{x}}-\frac{\cos{x}}{x},$$ which is easy.

Indeed, $$\lim_{x\rightarrow0^+}f'(x)=\lim_{x\rightarrow0^+}\left(-\frac{\cos{x}}{\sin^2x}-\frac{-x\sin{x}-\cos{x}}{x^2}\right)=$$ $$=\lim_{x\rightarrow0^+}\left(\frac{\sin{x}}{x}+\frac{\cos x(\sin^2x-x^2)}{x^2\sin^2x}\right)=$$ $$=1+\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}\lim_{x\rightarrow0^+}\frac{\frac{\sin{x}}{x}+1}{\frac{\sin^2x}{x^2}}=$$ $$=1+2\lim_{x\rightarrow0^+}\frac{\cos{x}-1}{3x^2}=1-\frac{1}{3}\lim_{x\rightarrow0^+}\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}=\frac{2}{3}>0$$ and we are done!

  • $\begingroup$ We need monotone to use Dirichlet, this is problem :( $\endgroup$ – Yury Gladkikh Oct 16 '17 at 17:07
  • $\begingroup$ @Yury Gladkikh Our problem is solved. Now tell something about Dirichlet and we are done! $\endgroup$ – Michael Rozenberg Oct 16 '17 at 17:10
  • $\begingroup$ I need to prove convergence of the series, not sequence( $\endgroup$ – Yury Gladkikh Oct 16 '17 at 17:17
  • $\begingroup$ So, for x this function increases in zero, and for n it will decrease in +inf, is it right? Thank you! $\endgroup$ – Yury Gladkikh Oct 16 '17 at 18:47
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    $\begingroup$ Yes, Yury, of course! You are welcome! $\endgroup$ – Michael Rozenberg Oct 16 '17 at 18:48

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