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Prove this inequality for all integers $m$:

$$∑_{n=2}^{m} \frac{1-n^{2α-1}}{n^{\alpha}} > \frac{1-(m+1)^{2α-1}}{(m+1)^{\alpha}}$$

for all $0<α<1/2$ and $m>2$.

DER
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    $$(1-n^{2\alpha-1})/n^\alpha=1/n^{\alpha} - n^{\alpha-1}$$ Now use the formula for geometric series. – J.R. Feb 18 '14 at 16:32
  • @TooOldForMath: But the summation is with respect to $n$ not to $α$ – DER Feb 18 '14 at 16:42
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    This is false in general. Here is an easy counterexample for $\alpha=0, m=3:$ The LHS is $(1-\frac{1}{2}) + (1-\frac{1}{3})= \frac{7}{6}$ and the RHS is $1-\frac{1}{4}=\frac{3}{4}$. Numerical computations show that for almost all $\alpha<1/2$ and $m>2$ the inequality should be reversed, i.e. the function $$ f(\alpha, m) = \frac{1-(m+1)^{2\alpha-1}} {(m+1)^{\alpha}} - \sum_{n=2}^m\frac{1-n^{2\alpha-1}}{n^\alpha}$$ is negative. As it stands it seems only valid for $\alpha<1/2$ and $m=2.$ – gammatester Feb 19 '14 at 09:46
  • @gammatester: It become true if I remove the value $α=0$, – DER Feb 19 '14 at 14:12
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    It can't be. The right hand side, $$\frac{1}{(m+1)^\alpha} - \frac{1}{(m+1)^{1-\alpha}},$$ tends to $0$ for $m\to\infty$, but the left hand side is positive and increasing. – Daniel Fischer Feb 19 '14 at 14:18
  • @DanielFischer: So the inequality should be reversed. – DER Feb 19 '14 at 14:24
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    No, reversing does not make it true: for $\alpha=0.25, m=2$ the LHS is $0.24629$ and the RHS is $0.32114,;$ see my first comment. – gammatester Feb 19 '14 at 14:29
  • @gammatester: Ok I have changed the question and I seeking a proof. – DER Feb 19 '14 at 14:40

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