자연계열 1-2번

[1.2] 삼각형 $\displaystyle \mathrm{Q}_1\mathrm{P}_2\mathrm{O}$은 각 $\displaystyle \mathrm{Q}_1\mathrm{P}_2\mathrm{O}$가 직각인 직각이등변삼각형이고 빗변의 길이가 $\displaystyle 1$이므로$\displaystyle \overline{\mathrm{Q}_1\mathrm{P}_2}=\overline{\mathrm{OP}_2}=\frac1{\sqrt2},\quad\overline{\mathrm{P}_1\mathrm{P}_2}=1-\frac1{\sqrt2}$이다. 따라서 $\displaystyle a_1=\frac{\sqrt2-1}4$이다. 모든 자연수 $\displaystyle n$에 대하여 삼각형 $\displaystyle \mathrm{Q}_n\mathrm{P}_{n+1}\mathrm{O}$이 각 $\displaystyle \mathrm{Q}_n\mathrm{P}_{n+1}\mathrm{O}$가 직각인 직각이등변삼각형이므로,$\displaystyle \overline{\mathrm{Q}_n\mathrm{P}_{n+1}}=\overline{\mathrm{OP}_{n+1}},\quad\overline{\mathrm{OP}_{n+1}}=\frac1{\sqrt2}\overline{\mathrm{OQ}_n}$이고, 문제의 조건에서 $\displaystyle \overline{\mathrm{OP}_n}=\overline{\mathrm{OQ}_n}$이다. 따라서$\displaystyle \overline{\mathrm{OP}_{n+1}}=\frac1{\sqrt2}\overline{\mathrm{OP}_n},\quad\overline{\mathrm{Q}_{n+1}\mathrm{P}_{n+2}}=\overline{\mathrm{OP}_{n+2}}=\frac1{\sqrt2}\overline{\mathrm{OP}_{n+1}}=\frac1{\sqrt2}\overline{\mathrm{Q}_n\mathrm{P}_{n+1}},$$\displaystyle a_n=\frac12\times\overline{\mathrm{P}_n\mathrm{P}_{n+1}}\times\overline{\mathrm{Q}_n\mathrm{P}_{n+1}}=\frac12\times(\overline{\mathrm{OP}_n}-\overline{\mathrm{OP}_{n+1}})\times\overline{\mathrm{Q}_n\mathrm{P}_{n+1}}$로부터$\displaystyle \begin{aligned}a_{n+1}&=\frac12\times(\overline{\mathrm{OP}_{n+1}}-\overline{\mathrm{OP}_{n+2}})\times\overline{\mathrm{Q}_{n+1}\mathrm{P}_{n+2}}\\&=\frac12\times\left(\frac1{\sqrt2}\overline{\mathrm{OP}_n}-\frac1{\sqrt2}\overline{\mathrm{OP}_{n+1}}\right)\times\frac1{\sqrt2}\times\overline{\mathrm{Q}_n\mathrm{P}_{n+1}}\\&=\frac14\times(\overline{\mathrm{OP}_n}-\overline{\mathrm{OP}_{n+1}})\times\overline{\mathrm{Q}_n\mathrm{P}_{n+1}}=\frac12a_n,\end{aligned}$즉 수열 $\displaystyle \{a_n\}$은 첫째항이 $\displaystyle \frac{\sqrt2-1}4$이고 공비가 $\displaystyle \frac12$인 등비수열이므로, 급수 $\displaystyle \sum_{n=1}^\infty a_n$의 합은$\displaystyle \frac{\frac{\sqrt2-1}4}{1-\frac12}=\frac{\sqrt2-1}2$이다.