등비급수와 부등식

$\displaystyle \sum\limits _{n = 1} ^{\infty} \left( \left| a _{n} \right| + a _{n} \right) = \frac{40}{3}$, $\displaystyle \sum\limits _{n = 1} ^{\infty} \left( \left| a _{n} \right| - a _{n} \right) = \frac{20}{3}$이므로 $\displaystyle \sum\limits _{n = 1} ^{\infty} \left| a _{n} \right| = 10$, $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n} = \frac{10}{3}$ 등비급수 $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n}$이 수렴하므로 등비수열 $\left\{ a _{n} \right\}$의 공비를 $r$라 하면 $- 1 < r < 1$ 수열 $\left\{ a _{n} \right\}$이 등비수열이면 $\left\{ \left| a _{n} \right| \right\}$도 등비수열이므로 등비수열 $\left\{ \left| a _{n} \right| \right\}$의 공비는 $\left| r \right|$이다. $\displaystyle \frac{\left| a _{1} \right|}{1 - \left| r \right|} = 10$, $\displaystyle \frac{a _{1}}{1 - r} = \frac{10}{3}$ (ⅰ) $0 \leq r < 1$일 때 $\left| a _{1} \right| = 10 - 10 r$, $\displaystyle a _{1} = \frac{10}{3} - \frac{10}{3} r$이므로 $\displaystyle 10 - 10 r = \frac{10}{3} - \frac{10}{3} r$ 또는 $\displaystyle 10 - 10 r = - \frac{10}{3} + \frac{10}{3} r$ $\therefore$ $r = 1$ 그런데 $0 \leq r < 1$이므로 조건을 만족시키지 않는다. (ⅱ) $- 1 < r < 0$일 때 $\left| a _{1} \right| = 10 + 10 r$, $\displaystyle a _{1} = \frac{10}{3} - \frac{10}{3} r$이므로 $\displaystyle 10 + 10 r = \frac{10}{3} - \frac{10}{3} r$ 또는 $\displaystyle 10 + 10 r = - \frac{10}{3} + \frac{10}{3} r$ $\displaystyle r = - \frac{1}{2} \left( \because - 1 < r < 0 \right)$, $a _{1} = 5$ (ⅰ), (ⅱ)에서 $a _{1} = 5$, $\displaystyle r = - \frac{1}{2}$ $\displaystyle \sum\limits _{k = 1} ^{2 n} \left\{ \left( - 1 \right) ^{\frac{k \left( k + 1 \right)}{2}} \times a _{m + k} \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{n} \left\{ \left( - 1 \right) ^{k \left( 2 k - 1 \right)} \times a _{m + 2 k - 1} + \left( - 1 \right) ^{k \left( 2 k + 1 \right)} \times a _{m + 2 k} \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{n} \left[ \left( - 1 \right) ^{k} \left\{ 5 \times \left( - \frac{1}{2} \right) ^{m + 2 k - 2} + 5 \times \left( - \frac{1}{2} \right) ^{m + 2 k - 1} \right\} \right]$ $\displaystyle = \sum\limits _{k = 1} ^{n} \left\{ \frac{5}{2} \times \left( - \frac{1}{2} \right) ^{m - 2} \times \left( - \frac{1}{4} \right) ^{k} \right\}$ 이므로 $\displaystyle \lim\limits _{n \rightarrow \infty} {\sum\limits _{k = 1} ^{2 n} \left\{ \left( - 1 \right) ^{\frac{k \left( k + 1 \right)}{2}} \times a _{m + k} \right\}}$ $\displaystyle = \sum\limits _{n = 1} ^{\infty} \left\{ \frac{5}{2} \times \left( - \frac{1}{2} \right) ^{m - 2} \times \left( - \frac{1}{4} \right) ^{n} \right\}$ $\displaystyle = \frac{- \frac{5}{8} \times \left( - \frac{1}{2} \right) ^{m - 2}}{1 - \left( - \frac{1}{4} \right)}$$\displaystyle = \left( - \frac{1}{2} \right) ^{m - 1}$ $\displaystyle \left( - \frac{1}{2} \right) ^{m - 1} > \frac{1}{700}$을 만족시키는 자연수 $m$은 홀수이고, $m$이 홀수이면 $\displaystyle \left( - \frac{1}{2} \right) ^{m - 1} = \frac{1}{2 ^{m - 1}} > \frac{1}{700}$ $2 ^{m - 1} < 700$ 이므로, 이를 만족시키는 홀수인 자연수 $m$의 값의 합은 $1 + 3 + 5 + 7 + 9 = 25$