Question

Which of these sequences are convergent? (Select all that apply) (A) An= _cos (2n) 5" g" (B) (n = 78 + 4 2n + (-1)"5 8n - (-1)"4 (D) «= (-2)" 2" + m3 (E) (n = 3 + 4" (F) (n = cos SAMSUNG

Answer 1

$\\$Answer A)$\\\\ a_n = \frac{\cos^4(2n)}{5^n}\\\\ $We know that $\\\\ -1 \leq \cos^4(2n) \leq 1 \\\\ \Rightarrow -\frac{1}{5^n} \leq \frac{\cos^4(2n)}{5^n} \leq \frac{1}{5^n}\\\\ \Rightarrow \lim_{n\to \infty }-\frac{1}{5^n} \leq \lim_{n\to \infty }\frac{\cos^4(2n)}{5^n} \leq\lim_{n\to \infty } \frac{1}{5^n}\\\\ \Rightarrow 0 \leq \lim_{n\to \infty }\frac{\cos^4(2n)}{5^n} \leq 0 \\\\ \Rightarrow \lim_{n\to \infty }\frac{\cos^4(2n)}{5^n} = 0\:\:\:\: $ by Sandwich Theorem$\\\\ $Hence, $a_n = \frac{\cos^4(2n)}{5^n} $ converges$\\\\$

$\\$Answer B)$ \\\\a_n = \frac{9^n}{n^8+4}\\\\$Clearly, $ a_n $ is a sequence of positive real numbers$\\\\ \lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \lim_{n\to \infty}\frac{9^{n+1}\:\big(n^8 + 4\big)}{\big((n+1)^8+4\big)\: 9^n} = \lim_{n\to \infty}\frac{9n^8\cdot9^{n}\:\big(1 + \frac{4}{n^8}\big)}{n^8\:\big((1+\frac1n)^8+\frac{4}{n^8}\big)\: 9^n}\\\\ \Rightarrow \lim_{n\to \infty} \frac{a_{n+1}}{a_n} =\lim_{n\to \infty}\frac{9\:\big(1 + \frac{4}{n^8}\big)}{\:\big((1+\frac1n)^8+\frac{4}{n^8}\big)} = \frac{9(1+0)}{(1+0)^8 + 0} = 9 > 1\\\\\\ $Hence, the series $a_n = \frac{9^n}{n^8+4} $ diverges by Ratio Test$$

$\\$Answer C)$ \\\\a_n = \frac{2n+ (-1)^n5}{8n-(-1)^n4}\\\\ \lim_{n\to \infty} a_n = \lim_{n\to \infty} \frac{2n+ (-1)^n5}{8n-(-1)^n4}= \lim_{n\to \infty} \frac{n\big(2+ \frac{(-1)^n5}{n}\big)}{n\big(8-\frac{(-1)^n4}{n}\big)}\\\\ \Rightarrow \lim_{n\to \infty} a_n = \lim_{n\to \infty} \frac{2+ \frac{(-1)^n5}{n}}{8-\frac{(-1)^n4}{n}} = \frac{2+0}{8-0} = \frac{1}{4}\\\\ $Hence, the series $a_n = \frac{2n+ (-1)^n5}{8n-(-1)^n4} $ converges$\\\\$

$\\$Answer D)$\\\\ a_n = (-2)^n\\\\ a_{2n} = (-2)^{2n} = 4^n \to \infty $ as $ n\to \infty \\\\$Hence, $ a_n $ has a sub-sequence $a_{2n} $ that diverges to $ \infty.\\\\ $ Hence, $ a_n = (-2)^n $ cannot converge.$$

$\\$Answer E)$ \\\\a_n = \frac{2^n + n^5 }{3^n + 4^n}\\\\ \lim_{n\to \infty} a_n = \lim_{n\to \infty} \frac{2^n + n^5 }{3^n + 4^n}= \lim_{n\to \infty} \frac{(\frac{2}{4})^n + ({n^5}/{4^n} )}{(\frac{3}{4})^n + 1^n}\\\\ $We know that, as $ n\to \infty, $ we have $ a^n \to 0 $ for $|a| < 1\\\\ $Hence, $\\\\ (\frac{2}{4})^n \to 0 $ as $ n \to \infty\\\\ (\frac{3}{4})^n \to 0 $ as $ n \to \infty\\\\ 1^n = 1 \to 1 $ as $ n \to \infty\\\\ $Let $ L = \lim_{n\to \infty} \frac{n^5}{4^n} \:\:\:\:(= \frac{\infty}{\infty} $ form$)\\\\ $Apply L'hopital Rule$\\\\ \Rightarrow L = \lim_{n\to \infty} \frac{5\:n^4}{4^n\: \ln 4} \:\:\:\: (= \frac{\infty}{\infty} $ form$)\\\\ $Apply L'hopital Rule again$\\\\ \Rightarrow L = \lim_{n\to \infty} \frac{20\:n^3}{4^n\: \ln^2 4} \:\:\:\:(= \frac{\infty}{\infty} $ form$)\\\\ $Apply L'hopital Rule again$\\\\ \Rightarrow L = \lim_{n\to \infty} \frac{60\:n^2}{4^n\: \ln^3 4} \:\:\:\:(= \frac{\infty}{\infty} $ form$)\\\\$

$\\$Apply L'hopital Rule again$\\\\ \Rightarrow L = \lim_{n\to \infty} \frac{120\:n}{4^n\: \ln^4 4} \:\:\:\:(= \frac{\infty}{\infty} $ form$)\\\\ $Apply L'hopital Rule$\\\\ \Rightarrow L = \lim_{n\to \infty} \frac{120}{4^n\: \ln^5 4} = 0\\\\ \Rightarrow \lim{n\to \infty} \frac{n^5}{4^n} = 0\\\\ $Combining all this, we get $\\\\ \lim_{n\to \infty }a_n= \lim_{n\to \infty} \frac{(\frac{2}{4})^n + ({n^5}/{4^n} )}{(\frac{3}{4})^n + 1^n}\\\\ = \frac{0 + 0}{0+1} = 0 \\\\ $Hence, the series $ a_n = \frac{2^n + n^5 }{3^n + 4^n} $ converges$$

$\\$Answer F)$\\\\ a_n = \cos(\frac{n\pi}{2})\\\\ a_{4n} = \cos(\frac{4n\pi}{2}) = \cos(2n\pi) = 1\\\\ \Rightarrow a_{4n} \to 1 $ as $n\to \infty \:\:\:\:---> $ (1)$\\\\ a_{4n+2} = \cos(\frac{(4n+2)\pi}{2}) = \cos((2n+1)\pi) = -1\\\\\Rightarrow a_{4n+2} \to (-1) $ as $ n\to \infty \:\:\:\:---> $ (2)$\\\\$From (1) and (2) we see that $ a_n $ has two sub-sequences $ a_{4n} $ and $ a_{4n+2} $ that converge to distinct limits.$\\\\ $Hence, the sequence $a_n = \cos(\frac{n\pi}{2}) $ is divergent$$

$\\\mathbf{Conclusion:}\\\\ \mathbf{Only\:the\:following\: sequences\:converge:}\\\\\\ a_n = \frac{\cos^4(2n)}{5^n}\\\\\\ a_n = \frac{2n+(-1)^n5}{8n-(-1)^n4}\\\\\\ a_n = \frac{2^n + n^5}{3^n + 4^n}$