Sign up ×
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 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!

share|cite|improve this question

1 Answer 1

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

share|cite|improve this answer

Your Answer


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.