Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)/Solution 1

Since $\sum_{n = 1}^{\infty} \frac{n a_{n} - 2 n + 1}{n + 1}$ converges, we have $\lim_{n \to \infty} \frac{n a_{n} - 2 n + 1}{n + 1} = 0$ . Hence:

$\begin{aligned} 0 & = \lim_{n \to \infty} \frac{n a_{n} - 2 n + 1}{n + 1} \\ = \lim_{n \to \infty} ((a_{n} - 2) \cdot \frac{n}{n + 1} + \frac{1}{n + 1}) \\ = \lim_{n \to \infty} (a_{n} - 2) \cdot \frac{n}{n + 1} + \lim_{n \to \infty} \frac{1}{n + 1} \\ = \lim_{n \to \infty} (a_{n} - 2) \cdot \frac{n}{n + 1} + 0 \\ = \lim_{n \to \infty} (a_{n} - 2) \cdot \lim_{n \to \infty} \frac{n}{n + 1} \\ = \lim_{n \to \infty} (a_{n} - 2) \cdot 1 \\ = \lim_{n \to \infty} (a_{n} - 2) \end{aligned}$

Therefore $\lim_{n \to \infty} (a_{n} - 2) = 0$ . Or rather, $\lim_{n \to \infty} a_{n} = 2$ .

On the other hand, $\ln (\frac{a_{n}}{a_{n + 1}}) = \ln a_{n} - \ln a_{n + 1}$ . Hence,

$\begin{aligned} - \ln a_{1} + \sum_{n = 1}^{k} \ln (\frac{a_{n}}{a_{n + 1}}) & = - \ln a_{1} + \sum_{n = 1}^{k} (\ln a_{n} - \ln a_{n + 1}) \\ = - \ln a_{1} + (\ln a_{1} - \ln a_{2}) + (\ln a_{2} - \ln a_{3}) + \dots + (\ln a_{k} - \ln a_{k + 1}) \\ = - \ln a_{k + 1} \end{aligned}$

Hence,

$\begin{aligned} - \ln a_{1} + \sum_{n = 1}^{\infty} \ln (\frac{a_{n}}{a_{n + 1}}) & = \lim_{k \to \infty} (- \ln a_{1} + \sum_{n = 1}^{k} \ln (\frac{a_{n}}{a_{n + 1}})) \\ = \lim_{k \to \infty} - \ln a_{k + 1} \\ = - \ln 2 . \end{aligned}$