1
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$

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.