Take the 2-minute tour ×
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.

I am trying to prove: For $n>1$

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

I am quite sure that this is correct. (checked with several arbitrary n values) But, no idea how to prove it. Any comments appreciated.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Consider that, for any $n\geq 2$, $f(x)=x^{n-1}$ is an increasing function on $\mathbb{R}^+$, so:

$$1^{n-1}+2^{n-1}+\ldots+(n-1)^{n-1}\leq\sum_{k=1}^{n-1}\int_{k}^{k+1}f(x)\,dx\leq\int_{0}^{n}x^{n-1}dx = n^{n-1}.$$

share|improve this answer
1  
It might be useful to add the details on why the initial inequality works (i.e., that each term $i^{n-1}$ underestimates the piece of the integral from $i$ to $i+1$, in standard Riemann box-sum fashion). –  Steven Stadnicki Nov 26 '12 at 16:48
    
Thanks a ton. Beautiful and easy proof. Can you also check the other question I posted? Proof should be similar to this one but does not trivially extend, I guess. –  Emre Nov 26 '12 at 21:37
    
@Emre: I gave a proof to your other inequality, where I used midpoint-convexity to strengthen this result. –  Jack D'Aurizio Nov 27 '12 at 9:30

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.