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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: $\displaystyle (c^{n+p1}+c^{n+p2}+...+c^{n})d < c^{n}\cdot \frac{1c^{p}}{1c}$ 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: 2,404 Thanks: 1306 
$\begin{align*} &(c^{n+p1} + c^{n+p2} + \dots + c^n) d = \\ \\ &c^n(c^{p1}+c^{p2}+\dots +1)d = \\ \\ &c^n d\displaystyle \sum \limits_{k=0}^{p1}~c^k = \\ \\ &c^n d \dfrac{1c^p}{1c} \end{align*}$ so you'll need some info on $d$ to complete the problem. 
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. 

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