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 Algebra Pre-Algebra and Basic Algebra Math Forum

 October 9th, 2019, 08:07 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Inequalities #3 Prove that $\displaystyle \: \displaystyle (1+e^{-1})\cdot (1+e^{-2}) \cdot ... \cdot (1+e^{-n}) < \displaystyle (e^{-1^2 }+e^{-2^2 }+e^{-3^2}+...+e^{-n^2 } )^-1} \; ,n >$ . e-euler constant . Method required ! Last edited by idontknow; October 9th, 2019 at 08:14 AM. October 9th, 2019, 09:07 AM #2 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle Using Bernoulli inequality: $L>1+e^{-1} + e^{-2}+...+e^{-n}$ Show that, $e^{-n}>e^{-n^2}$. Therefore, $L>1+e^{-1} + e^{-2}+...+e^{-n}>e^{-1^2}+e^{-2^2}+...+e^{-n^2}$ Invert the previous inequality, you get, $L<(e^{-1^2}+e^{-2^2}+...+e^{-n^2})^{-1}$ L stands for the left side of inequality. Thanks from idontknow Last edited by tahirimanov19; October 9th, 2019 at 09:12 AM. October 9th, 2019, 09:10 AM #3 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle $(1+x_1)(1+x_2)+...+(1+x_n) \ge 1+x_1+x_2+...+x_n$ $x_1,x_2,...,x_n>-1$ and $sgn(x_1)=sgn(x_2)=...=sgn(x_n)$ ------ Also, for any x>-1, $(1+x)^n \ge 1+nx, \; n \in \mathbb{N}, n>1$. And equality is true iff x=0. ------ Prove these. October 9th, 2019, 09:24 AM #4 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle If you want inequalities, I can post plenty of them... Thanks from idontknow Tags inequalities Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mj1395 Elementary Math 2 July 18th, 2016 08:10 PM Alexis87 Algebra 3 November 20th, 2013 04:57 AM drewm Algebra 1 July 2nd, 2011 07:10 PM outsos Real Analysis 15 January 1st, 2011 05:38 PM TungLHang Algebra 17 November 13th, 2010 04:44 AM

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