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Another inequality with powers: The proof for previous inequality does not trivially extend, I guess.

$$\text{For}\; n>2,\quad\quad(2n)^{n-1}> 1^{n-1}+3^{n-1}+...+\left( 2n-3\right) ^{n-1}+\left( 2n-1\right) ^{n-1}$$

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That's not a power series, it is a finite sum with constant powers. – Dennis Gulko Nov 26 '12 at 21:38
You are absolutely right, corrected. – Emre Nov 26 '12 at 21:39

If we divide both sides by $2^{n-1}$, we have to prove: $$\forall n>2,\quad n^{n-1} > \sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}.$$ Since $f(x)=x^{n-1}$ is a positive, continous, increasing and (midpoint-)convex function on $\mathbb{R}^+$, we have: $$ \left(k+\frac{1}{2}\right)^{n-1}< \int_{k}^{k+1}f(x)\,dx,$$ so: $$ \sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}<\int_{0}^{n}x^{n-1}\,dx=n^{n-1},$$ QED.

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thank you for your answer. I had employed a similar approach but stuck with your second inequality: Integral from k to k+1 being larger than the value at the midpoint. I believe it to be true but what would be the formal proof for that? – Emre Nov 27 '12 at 15:22
By midpoint-convexity of $f$, $$\int_{k}^{k+1}f(x) = \int_{0}^{1/2} f(k+1/2+t)+f(k+1/2-t)\, dt > \int_{0}^{1/2} 2 f(k+1/2) dt = f(k+1/2).$$ – Jack D'Aurizio Nov 27 '12 at 15:32
Excellent, thanks. – Emre Nov 27 '12 at 15:58

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