합에서 급수

[풀이1] $\displaystyle \lim\limits _{n \rightarrow \infty} {} S _{n} = \lim\limits _{n \rightarrow \infty} {} \frac{6 n}{n + 1} = 6$ $\displaystyle \lim\limits _{n \rightarrow \infty} {} S _{n + 1} = \lim\limits _{n \rightarrow \infty} {} \frac{6 n + 6}{n + 2} = 6$ $\displaystyle \sum\limits _{n = 1} ^{\infty} ( a _{n} + a _{n + 1} ) = \lim\limits _{n \rightarrow \infty} {} \sum\limits _{k = 1} ^{n} ( a _{k} + a _{k + 1} )$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( \sum\limits _{k = 1} ^{n} a _{k} + \sum\limits _{k = 1} ^{n} a _{k + 1} \right)$$\displaystyle = \lim\limits _{n \rightarrow \infty} {} \left( S _{n} + S _{n + 1} - a _{1} \right)$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {} S _{n} + \lim\limits _{n \rightarrow \infty} {} S _{n + 1} - a _{1}$ $= 6 + 6 - 3 = 9$ ($\because a _{1} = S _{1} = 3$) [풀이2] $a _{n} + a _{n+1} = \left( S _{n} - S _{n-1} \right) + \left( S _{n+1} - S _{n} \right)$ $=$$S _{n+1} - S _{n-1}$ $\displaystyle = \frac{6 ( n + 1 )}{n + 2} - {\frac{6 ( n - 1 )}{n}}$ $\displaystyle = \frac{6 n ( n + 1 ) - 6 ( n - 1 ) ( n + 2 )}{n ( n + 2 )}$ $\displaystyle = \frac{12}{n ( n + 2 )}$$\displaystyle = 6 \left( \frac{1}{n} - \frac{1}{n + 2} \right)$($n \geq 2$) $a _{1} + a _{2} = S _{2} = 4$이므로 $a _{n} + a _{n+1}$$\displaystyle = 6 \left( \frac{1}{n} - \frac{1}{n + 2} \right)$($n \geq 1$)