$\sum_{n\mathop=0}^\infty \frac{(-1)^n \pi^{2n}}{9^n(2n)!}$ I'm having trouble with finding the sum of the following series:
$$\sum_{n\mathop=0}^\infty \dfrac{(-1)^n \pi^{2n}}{9^n(2n)!}$$
Usually I would try to differentiate, but here its really inconvenient due to the factorial, which I'm not even sure it's going to help in any way. Any ideas?
 A: We need $$S=\sum_{n=0}^\infty\dfrac{(i\pi/3)^{2n}}{(2n)!}$$
Now as $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!},$
$$2S=e^{i\pi/3}+e^{-i\pi/3}=2\cos\dfrac\pi3$$
A: Using the Taylor series of cosine:
$$\begin{array}{rcl}
\displaystyle \sum_{n\mathop=0}^\infty \dfrac{(-1)^n \pi^{2n}}{9^n(2n)!}
&=& \displaystyle \sum_{n\mathop=0}^\infty \dfrac{(-1)^n \left(\frac\pi3\right)^{2n}}{(2n)!} \\
&=& \displaystyle \cos \left( \dfrac \pi 3 \right) \\
&=& \displaystyle \dfrac12 \\
\end{array}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\sum_{n = 0}^{\infty}{\pars{-1}^{n}\pi^{2n} \over 9^{n}\pars{2n}!} & =
\sum_{n = 0}^{\infty}{\ic^{2n}\pi^{2n} \over 3^{2n}\pars{2n}!} =
\sum_{n = 0}^{\infty}{\ic^{n}\pi^{n} \over 3^{n}\,n!}\,
{1 + \pars{-1}^{n} \over 2} =
{1 \over 2}\bracks{\sum_{n = 0}^{\infty}{\pars{\pi\ic/3}^{n} \over n!} +
\sum_{n = 0}^{\infty}{\pars{-\pi\ic/3}^{n} \over n!}}
\\[5mm] & =
\Re\sum_{n = 0}^{\infty}{\pars{\pi\ic/3}^{n} \over n!} =
\Re\exp\pars{{\pi \over 3}\,\ic} = \cos\pars{\pi \over 3} = \bbx{1 \over 2}
\end{align}
