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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.

Could anyone please help me get a direction.

Thank you very much for your help.

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    $\begingroup$ $\sum_{n=1}^\infty |\sin^{2n}x|$ is not bounded by $1/|\sin x|$. Take $x=\pi/2$ for example. $\endgroup$ Commented Jan 26, 2015 at 15:51
  • $\begingroup$ To apply leibniz you need to focus on $b_n=\dfrac{2^n \sin ^{2n}x }{n }$. $\endgroup$
    – user169373
    Commented Jan 26, 2015 at 15:53
  • $\begingroup$ @TimRaczkowski Thanks. Could you tell me how to apply the Abel's test here? $\endgroup$
    – MathMan
    Commented Jan 26, 2015 at 15:54
  • $\begingroup$ @Gato could you please tell me how to find the limiting value of $b_n$ in order to be able to apply Leibiniz $\endgroup$
    – MathMan
    Commented Jan 26, 2015 at 15:55

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much. Is it possible to apply the Abel's or Dirichlet's test here as well? $\endgroup$
    – MathMan
    Commented Jan 26, 2015 at 15:55
  • $\begingroup$ @kobe Correct me if I'm wrong, but $2\sin^2x=1-\cos2x$ therefore $0<2\sin^2x<2$. $\endgroup$ Commented Jan 26, 2015 at 15:58
  • $\begingroup$ @kobe Edited my comment. $\endgroup$ Commented Jan 26, 2015 at 16:02
  • $\begingroup$ @Tim I assumed $|\sin x| < \frac{1}{\sqrt{2}}$. $\endgroup$
    – kobe
    Commented Jan 26, 2015 at 16:03
  • $\begingroup$ @kobe Ah, I see know. :) $\endgroup$ Commented Jan 26, 2015 at 16:04
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We have $$\sum_{r=1}^\infty\frac{\left(-2\sin^2x\right)^n}n=-\ln[1-(-2\sin^2x)]$$

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

See Taylor series for $\log(1+x)$ and its convergence and Convergence for log 2

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  • $\begingroup$ Thanks. Could you please tell me if it's possible to apply the Dirichlets or the Abels test here? $\endgroup$
    – MathMan
    Commented Jan 26, 2015 at 15:56

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