Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I show that:

$$ a^n-1 \geq n\left(a^{\frac{n+1}{2}}-a^\frac{n-1}{2}\right)$$

$$ \sum_{k=0}^{n-1} a^k \geq na^\frac{n-1}{2}$$

$$ a>1, n\in\mathbb{N} $$

without studying the function

$$ f(x)=x^n-1 - n\left(x^{\frac{n+1}{2}}-x^\frac{n-1}{2}\right)$$?

share|cite|improve this question
up vote 6 down vote accepted

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+..+a+1)$$

By AM-GM

$$a^{n-1}+a^{n-2}+..+a+1 > na^{\frac{n-1}{2}}$$

Thus

$$a^n-1 >(a-1)na^{\frac{n-1}{2}}$$

share|cite|improve this answer
    
Thanks for noting me that. I misread the title. :) – S. Snape Aug 28 '12 at 13:17
    
Thank you for your answer! – Chon Aug 28 '12 at 16:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.