Compare Expression (Normal)

Algebra Level 4

Let S = 1 1 + 1 2 + 1 3 + + 1 2023 + 1 2024 S=\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 3 } } +\cdots+\frac { 1 }{ \sqrt { 2023 } } +\frac { 1 }{ \sqrt { 2024 } } . Without using calculator, compare S S with 88.

S > 88 S>88 S < 88 S<88 S = 88 S=88

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2 solutions

S 2 = 1 2 1 + 1 2 2 + 1 2 3 + . . . + 1 2 2023 + 1 2 2024 H a v e : 0 < 2 n < n + n + 1 1 2 n > 1 n + n + 1 = n + n + 1 S 2 > 1 + 2 2 + 3 . . . 2023 + 2024 2024 + 2025 = 1 + 2025 = 1 + 45 = 44 S > 88. \frac { S }{ 2 } =\frac { 1 }{ 2\sqrt { 1 } } +\frac { 1 }{ 2\sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 3 } } +...+\frac { 1 }{ 2\sqrt { 2023 } } +\frac { 1 }{ 2\sqrt { 2024 } } \\ Have:\quad 0<2\sqrt { n } <\sqrt { n } +\sqrt { n+1 } \Rightarrow \frac { 1 }{ 2\sqrt { n } } >\frac { 1 }{ \sqrt { n } +\sqrt { n+1 } } =-\sqrt { n } +\sqrt { n+1 } \\ \Rightarrow \frac { S }{ 2 } >-\sqrt { 1 } +\sqrt { 2 } -\sqrt { 2 } +\sqrt { 3 } -...-\sqrt { 2023 } +\sqrt { 2024 } -\sqrt { 2024 } +\sqrt { 2025 } =-\sqrt { 1 } +\sqrt { 2025 } =-1+45=44\\ \Rightarrow S>88.

Otto Bretscher
Dec 16, 2015

Consider the left Riemann sum of f ( x ) = 1 x f(x)=\frac{1}{\sqrt{x}} on [ 1 , 2025 ] [1,2025] with mesh size 1. Now 1 2025 x 1 / 2 d x = 2 [ x 1 / 2 ] 1 2025 = 2 ( 45 1 ) = 88 < k = 1 2024 1 k = S \int_{1}^{2025}x^{-1/2}dx=2[x^{1/2}]_{1}^{2025}=2(45-1)=88<\sum_{k=1}^{2024}\frac{1}{\sqrt{k}}=S

Cool, thanks :D

Tôn Ngọc Minh Quân - 5 years, 5 months ago

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