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Mathematics
General Math
Math Proof Training and Practice
Math Challenge - December 2021
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[QUOTE="julian, post: 6574855, member: 142346"] 10. We have \begin{align*} \prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right) & = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( 1 - \frac{1}{n} \right) \nonumber \\ & = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( \frac{n-1}{n} \right) \nonumber \\ & = \lim_{m \rightarrow \infty} \frac{1}{m} = 0 \end{align*} Hence the product \begin{align*} \prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right) \end{align*} converges. Now we examine the convergence of \begin{align*} \prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) \end{align*} We need a lemma: If a sequence ##(\alpha_n)_{n=1}^\infty## satisfies ##\sum_{n=1}^\infty |1 - \alpha_n| \leq \frac{1}{2}##, then for all ##m \geq 1## we have \begin{align*} \left| 1 - \prod_{n=1}^m \alpha_n \right| \leq 2 \sum_{n=1}^m |1 - \alpha_n| \end{align*} Which is easily proved by induction: The base case ##m = 1## is obvious, so suppose the inequality holds for some ##m \geq 2##. By the inductive hypothesis we have \begin{align*} \left| \prod_{n=1}^m \alpha_n \right| & \leq 1 + \left| 1 - \prod_{n=1}^m \alpha_n \right| \nonumber \\ & \leq 1 + 2 \sum_{n=1}^m |1 - \alpha_n| \leq 2 \end{align*} We use this in: \begin{align*} \left| 1 - \prod_{n=1}^{m+1} \alpha_n \right| & = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n - \prod_{n=1}^{m+1} \alpha_n \right| \nonumber \\ & = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n (1 - \alpha_{m+1}) \right| \nonumber \\ & \leq \left| 1 - \prod_{n=1}^m \alpha_n \right| + \left| \prod_{n=1}^m \alpha_n \right| \cdot |1 - \alpha_{m+1}| \nonumber \\ & \leq 2 \sum_{n=1}^m |1 - \alpha_n| + 2 |1 - \alpha_{m+1}| \nonumber \\ & = 2 \sum_{n=1}^{m+1} |1 - \alpha_n| \end{align*} where we have used the inductive hypothesis again. Note that in particular that ##|1 - \prod_{n=1}^m \alpha_n| \leq 1## holds for all ##m \geq 1##. We now show that there exists an ##N \geq 3## such that ##\sum_{n=N}^\infty \left| \frac{4}{n^2} \right| \leq \frac{1}{2}##. We do this by proving the series ##\sum_{n=3}^\infty \frac{4}{n^2}## convergences. Note \begin{align*} 0 & \leq \frac{4}{n^2} \leq \frac{4}{n^2 - n} \quad \text{for } n \geq 2 \end{align*} and \begin{align*} \sum_{n=3}^m \frac{4}{n^2 - n} & \leq 4 \sum_{n=3}^m \left( \frac{1}{n-1} - \frac{1}{n} \right) \nonumber \\ & = 4 \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{m-1} - \frac{1}{m} \right) \right] \nonumber \\ & = 2 - 4 \frac{1}{m} \end{align*} implying ##\lim_{m \rightarrow \infty} \sum_{n=3}^m \frac{4}{n^2 - n} = 2##, which in turn implies that ##\sum_{n=3}^\infty \frac{4}{n^2}## converges. As ##\sum_{n=3}^\infty \frac{4}{n^2}## converges there exists an ##N \geq 3## such that \begin{align*} \sum_{n=N}^\infty \frac{4}{n^2} & \leq \frac{1}{2} \end{align*} (In fact ##\sum_{n=9}^\infty \frac{4}{n^2} < \frac{1}{2}##). We can now use the above lemma: As the sequence ##\left( 1 - \frac{4}{n^2} \right)_{n=N}^\infty## satisfies ##\sum_{n=N}^\infty \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq \frac{1}{2}##, then for all ##m \geq N## we have \begin{align*} \left| 1 - \prod_{n=N}^m \left( 1 - \frac{4}{n^2} \right) \right| \leq 2 \sum_{n=N}^m \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq 1 \end{align*} So that \begin{align*} \left| 1 - \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right) \right| \leq 1 \end{align*} Hence \begin{align*} \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right) \end{align*} convergences, and given that ##\prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right)## is finite for finite ##N##, we finally have that the product \begin{align*} \prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right) \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right) \end{align*} converges. [/QUOTE]
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Math Proof Training and Practice
Math Challenge - December 2021
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