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Evaluate $\sum_{0}^{\infty}\frac{(-1)^n\pi^{2n+1}}{3^{2n}(2n+1)!}$

For this homework problem, I really don't have any clue how to start it, so any hints are welcome. But my first intuition would be to determine its convergence using the ratio test.

But even if I did determine that it was convergent by RT, I'm not aware of how to evaluate it for the sum.

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Hint: What is the Taylor series of the sine function centered at $0$?

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  • $\begingroup$ Well, the Taylor series of sin x is $\sum_{0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$. So the x value is then replaced by $\pi$, so its $\sum_{0}^{\infty}(-1)^n\frac{\pi^{2n+1}}{(2n+1)!}$. $\endgroup$ – Christian Martinez Apr 14 at 17:23
  • $\begingroup$ Why do you take $x=\pi$? Take $x=\frac\pi3$ instead. $\endgroup$ – José Carlos Santos Apr 14 at 17:25
  • $\begingroup$ So then, there is just the $\frac{1}{3^{2n}}$ portion, and then I evaluate that at 0? So would it be $\frac{1}{3^{2n}}*sin(x)$ at a = 0 --> $\frac{1}{3}*0$? $\endgroup$ – Christian Martinez Apr 14 at 17:26
  • $\begingroup$ Oh, I see how $x = \frac{\pi}{3}$. So then would because the Maclaurin series converges to $\frac{1}{1^{2n}}sin(\frac{\pi}{3})$ $\endgroup$ – Christian Martinez Apr 14 at 17:28
  • $\begingroup$ Note that your series is equal to $3\sum_{n=0}^\infty\frac{(-1)^n\left(\frac\pi3\right)^n}{(2n+1)!}$. So, the sum is $3\sin\left(\frac\pi3\right)=\frac{3\sqrt3}2$. $\endgroup$ – José Carlos Santos Apr 14 at 17:29
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$$\dfrac{(-1)^n\pi^{2n+1}}{3^{2n}}=-3i\left(\dfrac{i\pi}3\right)^{2n+1}$$

So, the required sum $$S=-3i\dfrac{e^{i\pi/3}-e^{-i\pi/3}}2$$

using https://proofwiki.org/wiki/Power_Series_Expansion_for_Exponential_Function

$$\implies S=-3i\cdot i\sin\dfrac\pi3$$

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If you write down a few terms you will see that your series is indeed the Taylor series of $3\sin x$ evaluated at $x=\pi/3$

Thus the answer is $3\sin(\pi/3)$

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