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 July 11th, 2018, 03:35 PM #1 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Why is the following inequality correct? Why is the following correct: $\displaystyle (c^{n+p-1}+c^{n+p-2}+...+c^{n})d < c^{n}\cdot \frac{1-c^{p}}{1-c}$ where $\displaystyle d=|a_{2}-a_{1}|$ (of series $\displaystyle a_{n}$) **The series $\displaystyle a_{n}$ is part of the problem, but I don't think it has anything to do with this specific inequality, so I didn't provide more details. Only this specific inequality (How to get from left side to right side?) July 11th, 2018, 03:52 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 \begin{align*} &(c^{n+p-1} + c^{n+p-2} + \dots + c^n) d = \\ \\ &c^n(c^{p-1}+c^{p-2}+\dots +1)d = \\ \\ &c^n d\displaystyle \sum \limits_{k=0}^{p-1}~c^k = \\ \\ &c^n d \dfrac{1-c^p}{1-c} \end{align*} so you'll need some info on $d$ to complete the problem. Thanks from Country Boy and Mathmatizer July 12th, 2018, 08:24 AM #3 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Thank you! $\displaystyle d=|a_{2}-a_{1}|$ and the way the series is defined makes $\displaystyle d = 0.5$ ($\displaystyle a_{2} = 1.5$ and $\displaystyle a_{1}= 1)$ so d<1 that's why the inequality is correct! Thank you! Last edited by Mathmatizer; July 12th, 2018 at 08:26 AM. Tags correct, inequality 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 StillAlive Calculus 5 September 2nd, 2016 11:45 PM OriaG Calculus 2 January 29th, 2013 09:14 AM cris(c) Algebra 1 July 19th, 2012 12:46 PM balste Advanced Statistics 2 September 3rd, 2009 11:20 AM cris(c) Calculus 0 December 31st, 1969 04:00 PM

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