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Therefore the series converges." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["23.", "Subsubsection"], Cell[TextData[{ "The series ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(n = 1\)\%\[Infinity] n\ \(sin(1\/n)\)\)]], " diverges because the terms do not approach 0: ", Cell[BoxData[ \(TraditionalForm \`lim\_\(n \[Rule] \[Infinity]\ \)n\ \(sin(1\/n)\) = \(lim\_\(x \[Rule] \[Infinity]\ \)\(sin(1\/x)\)\/\(1\/x\) = \ 1 \)\)]], " (by L'Hopital's Rule)." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["31.", "Subsubsection"], Cell[TextData[{ "For what values of ", Cell[BoxData[ \(TraditionalForm\`a\)]], " does the series ", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(n = 1\)\%\[Infinity]\((a\/\(n + 2\) - 1\/\(n + 4\))\)\)]], " converge?\n\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(n = 1\)\%\[Infinity]\((a\/\(n + 2\) - 1\/\(n + 4\))\)\ = \ \[Sum]\+\(n = 1\)\%\(\[Infinity]\ \)\(\((a - 1)\)\ n\ + \ 4\ a\ - \ 2\)\/\(\((n + 2)\) \((n + 4)\)\)\)]], "\n\nIf ", Cell[BoxData[ \(TraditionalForm\`a = 1\)]], ", then the series becomes ", Cell[BoxData[ \(TraditionalForm\`\(\ \[Sum]\+\(n = 1\)\%\(\[Infinity]\ \)2 \/\(\((n + 2)\) \((n + 4)\)\)\)\)]], ". Apply the integral test:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_1\%b\( 2\/\(\((x + 2)\) \((x + 4)\)\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(Log[5\/3] + Log[2 + b] - Log[4 + b]\)], "Output"] }, Open ]], Cell[TextData[{ "The limit ", Cell[BoxData[ \(TraditionalForm\`ln(\(b + 2\)\/\(b + 4\)) \[Rule] 1\)]], " as ", Cell[BoxData[ \(TraditionalForm\`b \[Rule] \[Infinity]\)]], " by L'Hopital's Rule, so the series converges.\n\nIf ", Cell[BoxData[ \(TraditionalForm\`a \[NotEqual] 0\)]], ", then the integral is" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_1\%b \(\(\((a - 1)\)\ x\ + \ 4\ a\ - \ 2\)\/\(\((x + 2)\) \((x + 4)\)\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-a\)\ Log[3] + Log[5] + a\ Log[2 + b] - Log[4 + b]\)], "Output"] }, Open ]], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`a\ > \ 0\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`\((b + 2)\)\^a\/\(b + 4\) \[Rule] \[Infinity]\)]], " and ", Cell[BoxData[ \(TraditionalForm \`ln(\((b + 2)\)\^a\/\(b + 4\)) \[Rule] \[Infinity]\)]], " as ", Cell[BoxData[ \(TraditionalForm\`b \[Rule] \[Infinity]\)]], ", and if ", Cell[BoxData[ \(TraditionalForm\`a\ < \ 0\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`\((b + 2)\)\^a\/\(b + 4\) \[Rule] 0\)]], " and ", Cell[BoxData[ \(TraditionalForm \`ln(\((b + 2)\)\^a\/\(b + 4\)) \[Rule] \(-\[Infinity]\)\)]], " as ", Cell[BoxData[ \(TraditionalForm\`b \[Rule] \[Infinity]\)]], ". Thus the series diverges when ", Cell[BoxData[ \(TraditionalForm\`a \[NotEqual] 0\)]], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["41.", "Subsubsection"], Cell[TextData[{ "a) From Exercise 33 we know that ", Cell[BoxData[ \(TraditionalForm \`\(ln(n + 1)\ \[LessEqual] 1 + 1\/2 + ... \) + 1\/n \[LessEqual] 1 + ln(n)\)]], ". This implies that ", Cell[BoxData[ \(TraditionalForm \`\(0 < ln(n + 1) - ln(n) \[LessEqual] 1 + 1\/2 + ... \) + 1\/n - ln(n) \[LessEqual] 1\)]], ", so the sequence ", Cell[BoxData[ \(TraditionalForm\`a\_n = \(1 + 1\/2 + ... \) + 1\/n - ln(n)\)]], "is bounded above and below.\n\nb) The graph of ", Cell[BoxData[ \(TraditionalForm\`y = 1\/x\)]], "for ", Cell[BoxData[ \(TraditionalForm\`n \[LessEqual] x \[LessEqual] n + 1\)]], "lies above the line ", Cell[BoxData[ \(TraditionalForm\`y = 1\/\(n + 1\)\)]], ", so ", Cell[BoxData[ \(TraditionalForm \`1\/\(n + 1\) < \[Integral]\_n\%\(n + 1\)\(1\/x\) \[DifferentialD]x\ = \ ln(n + 1) - ln(n)\)]], ". 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