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

Question

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.

Answer 1

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

Answer 2

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 $3\sin(\pi/3)$

Answer 3

$$\dfrac{(-1)^n\pi^{2n+1}}{3^{2n}}=-3i\left(\dfrac{i\pi}3\right)^{2n+1}$$

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

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