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Can anyone provide a slick proof of the following?

Let $0 < x \le 1$. Then $\displaystyle \sum_{k=0}^{n-1} x^k \ge \frac {1} {1 - (1 - 1/n)x}$.

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1 Answer 1

up vote 3 down vote accepted

As $1-(1-1/n)x>0$, we have to show that $(1-x)\sum_{k=0}^{n-1}x^k+\frac 1n\sum_{k=0}^{n-1}x^{k+1}\geq 1$. This is equivalent to $$1-x^n+\frac 1n\sum_{j=1}^nx^j\geq 1,$$ i.e., $$nx^n\leq \sum_{j=1}^nx^j.$$ This is true, as each terms in the sum is $\geq x^n$.

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Thanks--very elegant. –  Richard Hevener Sep 30 '12 at 20:48

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