[문제]
수열 \(\{a_n\}\) 의 첫째항부터 제 \(n\) 항까지의 합을 \(S_n\) 이라 할 때, \(S_n = \dfrac{6n}{n+1}\) 이다. \(\sum\limits_{n=1}^{\infty} (a_n+a_{n+1})\) 의 값은?
[풀이]
\[\begin{align*}a_n = S_n - S_{n-1}&=\dfrac{6n}{n+1}-\dfrac{6(n-1)}{n} \\ &= \dfrac{6n^2-6(n-1)(n+1)}{n(n+1)} = \dfrac{6n^2-(6n^2-6)}{n(n+1)} \\ &= \dfrac{6}{n(n+1)} = 6 \left( \dfrac{1}{n} - \dfrac{1}{n+1} \right)\end{align*}\]
\[\begin{align*}\sum\limits_{n=1}^{\infty} (a_n + a_{n+1}) &= \lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} (a_n + a_{n+1}) \\&= \lim_{n\rightarrow\infty} \left( \sum\limits_{k=1}^{n} a_n + \sum\limits_{k=1}^{n} a_{n+1} \right)\\&= \lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_n + \lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_{n+1}\end{align*}\]
\[\begin{align*}\lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_n &= \lim_{n\rightarrow\infty} \left( a_1 + a_2 + \cdots + a_n \right) \\ &= \lim_{n\rightarrow\infty} \left\{ 6 \left( \dfrac{1}{1} - \dfrac{1}{2} \right) + 6 \left( \dfrac{1}{2} - \dfrac{1}{3} \right) + \cdots + 6 \left( \dfrac{1}{n} - \dfrac{1}{n+1} \right) \right\} \\ &= \require{cancel}\lim_{n\rightarrow\infty} 6\left\{ \left( \dfrac{1}{1} - \cancel{\dfrac{1}{2}} \right) + \left(\cancel{\dfrac{1}{2}} - \cancel{ \dfrac{1}{3}} \right) + \cdots + \left( \cancel{\dfrac{1}{n}} - \dfrac{1}{n+1} \right) \right\} \require{cancel} \\ &= \lim_{n\rightarrow\infty} 6\left( \dfrac{1}{1} - \dfrac{1}{n+1} \right) = 6\end{align*}\]
\[\begin{align*}\lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_{n+1} &= \lim_{n\rightarrow\infty} \left( a_2 + a_3 + \cdots + a_{n+1} \right) \\ &= \lim_{n\rightarrow\infty} \left\{ 6 \left( \dfrac{1}{2} - \dfrac{1}{3} \right) + 6 \left( \dfrac{1}{3} - \dfrac{1}{4} \right) + \cdots + 6 \left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) \right\} \\ &= \require{cancel}\lim_{n\rightarrow\infty} 6\left\{ \left( \dfrac{1}{2} - \cancel{\dfrac{1}{3}} \right) + \left(\cancel{\dfrac{1}{3}} - \cancel{ \dfrac{1}{4}} \right) + \cdots + \left( \cancel{\dfrac{1}{n+1}} - \dfrac{1}{n+2} \right) \right\} \require{cancel} \\ &= \lim_{n\rightarrow\infty} 6\left( \dfrac{1}{2} - \dfrac{1}{n+2} \right) = 3\end{align*}\]
\[\therefore \lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_n + \lim_{n\rightarrow\infty} \sum\limits_{k=1}^{n} a_{n+1} =9\]