나머지정리와 급수

$a _{n} x ^{2} + a _{n} x + 2$를 $x - n$으로 나눈 나머지가 $25$이므로 $a _{n} n ^{2} + a _{n} n + 2 = 25$ $n ( n + 1 ) a _{n} = 23$ $\displaystyle \therefore a _{n} = \frac{23}{n ( n + 1 )}$ $\displaystyle \therefore \sum\limits _{{n} = 1} ^{\infty} a _{n} = \sum\limits _{{n} = 1} ^{\infty} \frac{23}{n ( n + 1 )}$$\displaystyle = 23 \lim\limits _{n \rightarrow \infty} \sum\limits _{{k} = 1} ^{n} \frac{1}{k ( k + 1 )}$ 이때, $\displaystyle \sum\limits _{{k} = 1} ^{n} \frac{1}{k ( k + 1 )} = \sum\limits _{{k} = 1} ^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$ $\displaystyle = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n + 1} \right)$ $\displaystyle = 1 - \frac{1}{n + 1}$ 이므로 $\displaystyle 23 \lim\limits _{n \rightarrow \infty} \sum\limits _{{k} = 1} ^{n} \frac{1}{k ( k + 1 )} = 23 \lim\limits _{n \rightarrow \infty} \left( 1 - \frac{1}{n + 1} \right)$ $= 23$