Does the series: $\sum\limits_{n=0}^{\infty }\left ( \tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n+1}{6n+1} \right )$ converge or diverge?

I've tried to apply Taylor expansion, but I wasn't able to find a proper way to apply it.

  • 1
    $\begingroup$ You might be able to show that $a_n=\frac{\pi n+1}{4n+2}\to\frac\pi4$ and $b_n=\frac{\pi n+1}{6n+1}\to\frac\pi6$, hence $\tan a_n\to1$, $\sin b_n\to\frac12$, and $\tan a_n-2\sin b_n\to0$. To go further, one needs estimating $x_n=a_n-\frac\pi4$ and $y_n=b_n-\frac\pi6$ to deduce estimates of $\tan a_n-1$ and $\sin b_n-\frac12$, can you do that? $\endgroup$ – Did Sep 7 '16 at 15:50
  • $\begingroup$ I think as opposed to your series converges $\,\sum\limits_{n=0}^\infty\left(\tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n(6-\pi)-3\sqrt{3}(\pi-2)}{6n(6-\pi)-3\sqrt{3}(\pi-2)}\right)\,$ (only for comparison). But I am not sure. $\endgroup$ – user90369 Sep 8 '16 at 8:09

We have $$ \tan\left(\frac{\pi}{4}+x\right)=1+2x+O(x^2),\qquad 2\sin\left(\frac{\pi}{6}+x\right)= 1+\sqrt{3}\,x+O(x^2)\tag{1}$$ $$ \frac{\pi n+1}{4n+2}=\frac{\pi}{4}+\left(\frac{1}{4}-\frac{\pi}{8}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{2}$$ $$ \frac{\pi n+1}{6n+1}=\frac{\pi}{6}+\left(\frac{1}{6}-\frac{\pi}{36}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{3}$$ hence by combining $(1),(2)$ and $(3)$ it follows that: $$ \tan\left(\frac{\pi n+1}{4n+2}\right)-2\sin\left(\frac{\pi n+1}{6n+1}\right) = \frac{C}{n}+O\left(\frac{1}{n^2}\right)\tag{4} $$ with $C\approx -0.42292335$, so the given series is divergent by comparison with the harmonic series.


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