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If \( S_{1}=1 \) and \( S_{n+1}=\frac{3 S_{n}+4}{3+2 S_{n}}, n \geq 1 \) Then \( \frac{\lim _{n \rightarrow \infty} S_{n}}{\lim _{n \rightarrow \infty}\left((2)^{n / 2}+\left(\frac{\pi}{4}\right)^{n}\right)^{\frac{1}{n}}} \) is (1) 1 (2) 2 (3) 4 (4) \( \frac{1}{2} \) |
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Answer» Correct option is (1) 1 \(S_1 = 1\) \(S_2 = \frac{3S_1 + 4}{3 + 2s_1} = \frac{3 +4}{3 + 2} = \frac 75\simeq1.4\) \(S_3 = \frac {3S_2 + 4}{3 + 2S_2} = \cfrac{3\times \frac 75 + 4}{3 + 2\times \frac 75} = \cfrac{\frac{21 + 20}5}{\frac{15 + 14}5} = \frac{41}{29}\simeq1.413\) \(S_4 = \frac{3S_3 + 4}{3 + 2S_3} = \cfrac{3\times \frac{41}{25} + 4}{3+ 2 \times \frac{41}{25}} = \cfrac{\frac{123 + 116}{29}}{\frac{87 + 82}{29}} = \frac{239}{169} \simeq 1.414\) \(S_5 = \frac{3S_4 + 4}{3 + 2S_4} = \cfrac{3\times \frac{239}{165}+4}{3+2\times \frac{235}{169}}= \cfrac{\frac{717+676}{169}}{\frac{507 + 478}{169}} = \frac{1393}{985} \simeq1.414213\) \(\lim\limits_{n \to \infty} S_n = \sqrt 2\) \(\lim\limits_{n \to \infty} (2^{n/2} + (\frac \pi 4)^n)^\frac 1n\) \((\infty \; type)\) \(= Exp \left\{\lim\limits_{n\to \infty} \frac{log(2^{n/2} + (\frac\pi4)^n)}{n}\right\}\) \((\frac \infty \infty \; type)\) \(= Exp \left\{\lim\limits_{n\to \infty} \cfrac{2^{n/2}\frac{log2}2 + (\frac \pi4)^n log (\frac \pi4)}{2^{n/2} + \left(\frac \pi4\right)^n}\right\}\) \(= Exp\left\{ \lim\limits_{n\to \infty} \cfrac{\frac12 log2 + \left(\frac\pi4\right)^n log(\frac\pi 4)2^{-n/2}}{1+ (\frac \pi4)^n \,2^{-n/2}}\right\}\) \(= e^{\frac 12 log 2} = e^{log2^\frac12} = 2^\frac12 = \sqrt 2\) \(\therefore \frac{\lim\limits_{n\to \infty } S_n}{\lim\limits_{n\to \infty} \left(2^\frac n2 + (\frac \pi 4)^n\right)^\frac1n} = \frac{\sqrt 2}{\sqrt 2} = 1\) |
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