For which real numbers x is the series $$\sum^{\infty}_{n=0} x^n \tan \left(\frac {x}{2^n}\right)$$ convergent and how (i.e. absolutely/conditionally)?

I have proved that the series converges absolutely for $|x|<2$, since $$\left| \frac{a_{n+1}}{a_n} \right|=\frac{|x|}{2} \left|1-\tan ^{2}\frac{x}{2^{n+1}}\right| \rightarrow\frac{|x|}{2}$$ and using the ratio test. Now I need to discuss the part where $|x| \ge 2$.

  • 1
    $\begingroup$ If the summation starts with $n = 0$, you have to think of $x = \pm \frac{\pi}{2}$. For $\lvert x\rvert \geqslant 2$, do the terms tend to $0$? $\endgroup$ – Daniel Fischer Aug 11 '18 at 14:03
  • $\begingroup$ So for $|x| \ge 2$ we have $ |a_n| \ge 2^n \tan \frac{2}{2^n} = 2^n \frac {\sin 2^{1-n}}{\cos 2^{1-n}}$ which goes to 2 for $n\to \infty$ and so the series could not be convergent. $\endgroup$ – Kryštof Aug 11 '18 at 15:10
  • $\begingroup$ Yes, if you throw in a "for sufficiently large $n$". This inequality need not hold if $\lvert x\rvert/2^n \geqslant \pi/2$. $\endgroup$ – Daniel Fischer Aug 11 '18 at 15:24
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    $\begingroup$ Note that the ratio test already covers $|x| > 2$ (you have shown the limit is $>1$ here so it diverges). The only case not covered by the ratio test is $|x| = 2$. $\endgroup$ – Winther Aug 11 '18 at 19:06

If $|x|>2 $ then $$\lim_{n\to \infty } x^n \tan \frac{x}{2^n} =\lim_{n\to \infty } 2 (\frac{x}{2})^{n+1}\frac{ \tan \frac{x}{2^n}}{\frac{x}{2^n}} =\infty.$$ Therefore the series diverges


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