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I read a proof which uses the following inequality:

Let $\alpha>p>0$, then $$\sum_{n=j}^{\infty}n^{-\alpha/p}\leq j^{-\alpha/p}+\frac{p}{\alpha-p}j^{(p-\alpha)/p}$$

I think it for a while, but have no idea. Thanks!

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up vote 3 down vote accepted

Consider the lower sum of the function $f(x)=x^{-\alpha/p}$ on $[1,\infty)$ i.e. the sum of box areas using the infimum of the function $x^{-\alpha/p}$ in each subinterval of lenght 1 of $[1,\infty)$ . Then, one can easily see that

$$ \left( \sum_{n=j}^\infty n^{-\alpha/p} \right)-j^{-\alpha/p} \leq \int_j^\infty x^{-\alpha/p}\;dx. $$

The result follows immediately since $\int_j^\infty x^{-\alpha/p}\;dx = \frac{p}{\alpha - p} j^{(p-\alpha)/p}$.

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