# convergence of $\sum_{n=1}^{\infty} (-1)^n \frac{2^n \sin ^{2n}x }{n }$

Find values of $x$ for which the following series converges $$\sum_{n=1}^{\infty} (-1)^n \dfrac{2^n \sin ^{2n}x }{n }$$

Attempt:

(a) Check for Absolute Convergence

If we consider $\sum_{n=1}^{\infty} (-1)^n \dfrac{2^n \sin ^{2n}x }{n }$, then : $\sum_{n=1}^{\infty} \sin ^{2n}x$ is bounded and $\le \dfrac {1}{|\sin x|}$.

But $\lim_{n \rightarrow \infty} \dfrac {2^n}{n} = \infty$.

Hence, we can't apply neither the Abel's Test nor the Dirichlets test.

This series does not seem to be absolutely convergent.

(b) Checking for conditional convergence

If $b_n= (-1)^n \dfrac{2^n \sin ^{2n}x }{n }$, then $\lim_{n \rightarrow \infty} b_n =\infty$.

Hence, we can't even apply he leibnitz condition for the conditional convergence of the series.

Thank you very much for your help.

• $\sum_{n=1}^\infty |\sin^{2n}x|$ is not bounded by $1/|\sin x|$. Take $x=\pi/2$ for example. – Tim Raczkowski Jan 26 '15 at 15:51
• To apply leibniz you need to focus on $b_n=\dfrac{2^n \sin ^{2n}x }{n }$. – user169373 Jan 26 '15 at 15:53
• @TimRaczkowski Thanks. Could you tell me how to apply the Abel's test here? – MathMan Jan 26 '15 at 15:54
• @Gato could you please tell me how to find the limiting value of $b_n$ in order to be able to apply Leibiniz – MathMan Jan 26 '15 at 15:55

The series converges absolutely if $|\sin x| < \frac{1}{\sqrt{2}}$. Indeed, if $|\sin x| < \frac{1}{\sqrt{2}}$, then $$\lim_{n\to \infty} \left\lvert (-1)^n \frac{2^n\sin^{2n} x}{n}\right\rvert^{1/n} = \lim_{n\to \infty} \frac{2\sin^2 x}{n^{1/n}} = 2\sin^2 x < 1$$ Thus, by the root test, $\sum_{n = 1}^\infty (-1)^n \frac{2^n\sin^{2n} x}{n}$ converges absolutely.

• Thank you very much. Is it possible to apply the Abel's or Dirichlet's test here as well? – MathMan Jan 26 '15 at 15:55
• @kobe Correct me if I'm wrong, but $2\sin^2x=1-\cos2x$ therefore $0<2\sin^2x<2$. – Tim Raczkowski Jan 26 '15 at 15:58
• @kobe Edited my comment. – Tim Raczkowski Jan 26 '15 at 16:02
• @Tim I assumed $|\sin x| < \frac{1}{\sqrt{2}}$. – kobe Jan 26 '15 at 16:03
• @kobe Ah, I see know. :) – Tim Raczkowski Jan 26 '15 at 16:04

We have $$\sum_{r=1}^\infty\frac{\left(-2\sin^2x\right)^n}n=-\ln[1-(-2\sin^2x)]$$

as $\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$

• Thanks. Could you please tell me if it's possible to apply the Dirichlets or the Abels test here? – MathMan Jan 26 '15 at 15:56