Answer to Question #200823 in Calculus for OhMoe

Question #200823

Consider the telescoping series

"n"

"Sn = \u03a3" "[ ( 1 \/ (k+1) ) - ( 1 \/ k )" ]

"k=2"



Which of the options below are incorrect?




1) "Sn = [ (1-n) \/ ( 2n + 2) )" ]



2)

"n + 1"

"Sn" = - "\u03a3 [ 1 \/ k ( k - 1) ]"

"k = 3"


3)

"n"

"Sn = \ufeff\u03a3 [ 1 \/ k ( k + 1) ]"

"k = 2"



4)

"\u221e"

"Sn = \ufeff\u03a3 [ 1 \/ k ( k + 1) ]" = "1\/2"

"k = 2"


5)

"\u221e"

Sn = "\u03a3 [ 1 \/ k ( k - 1) ]" "= - 1\/2"

"k = 3"



1
Expert's answer
2021-06-08T04:15:59-0400

Given, the telescoping series

"S_n=\u03a3_{k=2}^{n}(\\frac{1}{k+1}-\\frac{1}{k})\\\\\n=(\\frac{1}{3}-\\frac{1}{2})+(\\frac{1}{4}-\\frac{1}{3})+(\\frac{1}{5}-\\frac{1}{4})\n+\u2022\u2022\u2022+(\\frac{1}{n+1}-\\frac{1}{n})\\\\\n=(\\frac{1}{n+1}-\\frac{1}{2})\\\\\n=\\frac{1-n}{2n+2}\\\\\n\\text{Thus, option (1) is correct.}\\\\\nS_n=\u03a3_{k=2}^{n}(\\frac{1}{k+1}-\\frac{1}{k})\\\\\n=\u03a3_{k=2}^{n}\\frac{-1}{k(k+1)}\\\\\n=-(\\frac{1}{2.3}+\\frac{1}{3.4}+\\frac{1}{4.5}+\\frac{1}{6.7}+\\frac{1}{7.8}\n+\u2022\u2022\u2022+\\frac{1}{n(n+1)})\\\\\n=-(\\frac{1}{3.2}+\\frac{1}{4.3}+\\frac{1}{5.4}+\\frac{1}{7.6}+\\frac{1}{8.7}\n+\u2022\u2022\u2022+\\frac{1}{(n+1).n})\\\\\n=\u03a3_{k=3}^{n+1}\\frac{-1}{k(k-1)}\\\\\n\\text{Thus, option (2) is correct.}"

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