無限級数 \(\textstyle\sum\limits _ {n=0}^{\infty} \left( \dfrac{1}{2} \right)^n \cos \dfrac{n \pi}{6}\) の和を求めよ.
【 解 答 】
\(\alpha = \cos \dfrac{\pi}{6} +i \sin \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} +\dfrac{i}{2}\) とおくと
\(\alpha^n = \cos \dfrac{n \pi}{6} +i \sin \dfrac{n \pi}{6}\) なので
\[
\left( \dfrac{1}{2} \right)^n \cos \dfrac{n \pi}{6} = \dfrac{1}{2} \left\{ \left( \dfrac{\alpha}{2} \right)^n +\left( \dfrac{\overline{\alpha}}{2} \right)^n \right\}
\]
したがって, 求める和 \(S\) は
\[
S = \displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{2} \left\{ \dfrac{1 -\left( \dfrac{\alpha}{2} \right)^{n+1}}{1 -\dfrac{\alpha}{2}}
+\dfrac{1 -\left( \dfrac{\overline{\alpha}}{2} \right)^{n+1}}{1 -\dfrac{\overline{\alpha}}{2}} \right\}
\]
\(\left| \dfrac{\alpha}{2} \right| = \left| \dfrac{\overline{\alpha}}{2} \right| = \dfrac{1}{2}\) なので
\[
\displaystyle\lim _ {n \rightarrow \infty} \left| \dfrac{\alpha}{2} \right|^{n+1} = \displaystyle\lim _ {n \rightarrow \infty} \left| \dfrac{\overline{\alpha}}{2} \right|^{n+1} = 0
\]
ゆえに
\[
\displaystyle\lim _ {n \rightarrow \infty} \left( \dfrac{\alpha}{2} \right)^{n+1} = \displaystyle\lim _ {n \rightarrow \infty} \left( \dfrac{\overline{\alpha}}{2} \right)^{n+1} = 0
\]
したがって
\[\begin{align}
S & = \dfrac{1}{2} \left( \dfrac{1}{1 -\dfrac{\alpha}{2}}
+\dfrac{1}{1 -\dfrac{\overline{\alpha}}{2}} \right) \\
& = \dfrac{1}{2 -\alpha} +\dfrac{1}{2 -\overline{\alpha}} \\
& = \dfrac{4 -\alpha -\overline{\alpha}}{| 2 -\alpha |^2}
\end{align}\]
ここで
\[\begin{align}
| 2 -\alpha |^2 & = \left( 2 -\dfrac{\sqrt{3}}{2} \right)^2 +\left( \dfrac{1}{2} \right)^2 \\
& = 4 +\dfrac{3}{4} -2\sqrt{3} +\dfrac{1}{4} \\
& = 5 -2 \sqrt{3} \ , \\
4 -\alpha -\overline{\alpha} & = 4 -\sqrt{3}
\end{align}\]
なので
\[\begin{align}
S & = \dfrac{4 -\sqrt{3}}{5 -2 \sqrt{3}} \\
& = \dfrac{\left( 4 -\sqrt{3} \right) \left( 5 +2 \sqrt{3} \right)}{13} \\
& = \underline{\dfrac{14 +3 \sqrt{3}}{13}}
\end{align}\]