"What is the infinite sum of the alternating series?"
$$\sum_{n=0}^{\infty} \frac{(-1)^n \, (5 \pi)^{2n+1}}{6^{2n+1} \, (2n+1)!}$$
This problem was given to me along with the $\cos(x)$ series, and I know that the answer is $\dfrac{1}{2}$, just not how to get there.
I have tried plugging in $5 \pi$ for $x$, but then from there, I don't know how to compensate for the $6^{2n+1}$ in the denominator to get the full sum.
All help is welcome! Thank you!
We can answer your question by re-writing your sum as $$\sum_{n=0}^{\infty} \frac {(-1)^n(\frac {5π} 6)^{2n+1}} {(2n+1)!}$$ by moving the $6^{2n+1}$ into the denominator into the $(5π)^{2n+1}$ in the numerator as they have the same exponent and then immediately noticing that this is the Mclaurin series for sin(x) evaluated at $x= \frac{5π}{6}$.
We know that $\frac{5π} {6} = π - \frac {π}{6}$, so using the identity $\sin(a-b) = \sin(a) \cos(b) - \cos(a) \sin(b)$ we can evaluate the sin($\frac{5π}{6}$) as follows: $$\sin(\frac{5π}{6}) = \sin(π - \frac {π}{6}) = \sin(π) \cos(\frac {π}{6}) - \cos(π) \sin(\frac {π}{6}) = 0 - (-1)(\frac {1}{2}) = \frac{1}{2}$$
Hope this helps!