Find the supremum and infimum of each of the following sequence a) {((n−1)/(2n))} b) {(((−)^n n)/(2n+1))} c){((1+(−)^n )/3)} d) {sin((nπ)/2)} e) {(1/n) − sin((nπ)/2)} f) {(1+(1/(2n)))cos((nπ)/3)} Help!


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Question Number 192738 by Mastermind last updated on 25/May/23
$Find the supremum and infimum of each of the following sequence a) {((n−1)/(2n))} b) {(((−)^n n)/(2n+1))} c){((1+(−)^n )/3)} d) {sin((nπ)/2)} e) {(1/n) − sin((nπ)/2)} f) {(1+(1/(2n)))cos((nπ)/3)} Help!$
$Find the supremum and infimum$ $of each of the following sequence$ $a) {\frac{n - 1}{2 n}} b) {\frac{{(-)}^{n} n}{2 n + 1}} c) {\frac{1 + {(-)}^{n}}{3}}$ $d) {\sin \frac{n π}{2}} e) {\frac{1}{n} - \sin \frac{n π}{2}}$ $f) {(1 + \frac{1}{2 n}) \cos \frac{n π}{3}}$ $Help!$
	Answered by Rajpurohith last updated on 26/May/23

	$a) 1 b) s u p = \frac{1}{2} a n d i n f = - \frac{1}{2}$ $c) s u p = \frac{2}{3} a n d i n f = \frac{1}{3}$ $d) s u p = 1 a n d i n f = - 1$ $e) s u p = 2 a n d i n f = 0$ $f) s u p = \frac{3}{2} a n d i n f = 1$
	Commented by Mastermind last updated on 26/May/23

	$Thank you but i need solution$ $how you got each$
	Commented by MM42 last updated on 27/May/23

	$s i r$ $y o u r a n s w e r s a r e i n c o r r e c t . p l e a s e p a y a t y e n t i o n$ $m y a n s w e r s$
	Answered by MM42 last updated on 27/May/23

	$a) 0, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, . . . \to \frac{1}{2} \Rightarrow i n f = 0 & s u p = \frac{1}{2}$ $b)$ $f o r t h e ‘ ‘ n = o d d^{″} - \frac{1}{3}, - \frac{3}{7}, - \frac{5}{11}, . . . \to i n f = - \frac{1}{2} & s u p = - \frac{1}{3} (i)$ $f o r t h e ‘ ‘ n = {e v e n}^{″} \frac{2}{5}, \frac{4}{9}, \frac{6}{13}, . . . \to i n f = \frac{2}{5} & s u p = \frac{1}{2} (i i)$ $(i), (i i) \Rightarrow i n f = - \frac{1}{2} & s u p = \frac{1}{2}$ $c)$ $f o r t h e ‘ ‘ n = o d d^{″} 0, 0, 0, . . . \to i n f = s u p = 0$ $f o r t h e ‘ ‘ n = e v e n^{″} \frac{2}{3}, \frac{2}{3}, \frac{2}{3}, . . . \to i n f = s u p = \frac{2}{3}$ $\Rightarrow i n f = 0 & s u p = \frac{2}{3}$ $d) 1, 0, - 1, 0, 1, 0, - 1, . . . \Rightarrow i n f = - 1 & s u p = 1$ $e)$ $f o r t h e ‘ ‘ n = {e v e n}^{″} e_{n} = \frac{1}{n} \Rightarrow \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, . . . \to 0 \Rightarrow i n f = 0 & s u p = \frac{1}{2} (i)$ $f o r t h e ‘ ‘ n = 4 k + 1^{″} e_{n} = \frac{1}{n} - 1 \Rightarrow 0, - \frac{4}{5}, - \frac{8}{9} . . . \to - 1 \Rightarrow i n f = - 1 & s u p = 0 (i i)$ $f o r t h e ‘ ‘ n = 4 k + 3^{″} e_{n} = \frac{1}{n} + 1 \Rightarrow \frac{4}{3}, \frac{8}{7}, \frac{12}{11} . . . \to 1 \Rightarrow i n f = 1 & s u p = \frac{4}{3} (i i i)$ $(i), (i i), (i i i) \Rightarrow i n f = - 1 & s u p = \frac{4}{3}$ $f)$ $f o r t h e ‘ ‘ n = 6 k^{″} f_{n} = 1 + \frac{1}{2 n} \Rightarrow \frac{13}{12}, \frac{25}{24}, \frac{37}{36}, . . . \to 1 \Rightarrow i n f = 1 & s u p = \frac{13}{12} (i)$ $f o r t h e ‘ ‘ n = 6 k - 3^{″} f_{n} = - (1 + \frac{1}{2 n}) \Rightarrow - \frac{7}{6}, - \frac{19}{18}, - \frac{31}{30}, . . . \to - 1 \Rightarrow i n f = - \frac{7}{6} & s u p = - 1 (i i)$ $f o r t h e ‘ ‘ n = 1, 5, 7, 11, 13, . .^{″} f_{n} = \frac{1}{2} (1 + \frac{1}{2 n}) \Rightarrow \frac{3}{4}, \frac{11}{20}, \frac{15}{28}, . . . \to \frac{1}{2} \Rightarrow i n f = \frac{1}{2} & s u p = \frac{3}{4} (i i i)$ $f o r t h e ‘ ‘ n = 2, 4, 8, 10, 14, . .^{″} f_{n} = - \frac{1}{2} (1 + \frac{1}{2 n}) \Rightarrow - \frac{5}{8}, - \frac{9}{16}, - \frac{17}{32}, - \frac{29}{56} . . . \to - \frac{1}{2} \Rightarrow i n f = - \frac{5}{8} & s u p = - \frac{1}{2} (i v)$ $(i), (i i), (i i i), (i v) \Rightarrow i n f = - \frac{7}{6} & s u p = \frac{13}{12}$
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