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Let $p\geq1$. Show that $$\frac{1}{(k+1)^p}\leq \int_k^{k+1}\frac{1}{x^p}dx\leq\frac{1}{k^p} $$ for every positive $k$. Hence show that $$\sum_{k=2}^{n+1}\frac{1}{k^p}\leq\int_1^{n+1}\frac{1}{x^p}dx\leq\sum_{k=1}^n\frac{1}{k^p}$$ for every positive integer $n$.

I turned $$\int_k^{k+1}\frac{1}{x^p}dx=\ln\left(\frac{k+1}{k}\right)$$for $p=1$. And $$\int_k^{k+1}\frac{1}{x^p}dx=\bigg[\left(\frac{k}{p-1}\right)\left(\frac{1}{k^p}\right)\bigg]-\bigg[\left(\frac{k}{p-1}\right)\left(\frac{1}{(k+1)^p}\right)\bigg]-\bigg[\left(\frac{1}{p-1}\right)\left(\frac{1}{(k+1)^p}\right)\bigg]$$for $p>1$.

But I don’t know how to compare it with $\frac{1}{(k+1)^p}$ and $\frac{1}{k^p}$.

Gary
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elizabeth10
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    well $$ \int_k^{k+1}\frac1{x^p} dx \ge \int_k^{k+1}\frac1{(k+1)^p} dx = \frac1{(k+1)^p} $$ – Chia Apr 08 '23 at 03:10
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    Other possible hint: $x^{-p}$ is decreasing on $(1, \infty)$. – Sean Roberson Apr 08 '23 at 03:15
  • The foundation of the comment of @Xier is that if $~h(x) \geq 0 ~: ~\forall ~x \in [a,b],~$ then $~\displaystyle \int_a^b h(x) dx \geq 0.~$ So, if $~f(x) \geq g(x) ~: ~\forall ~x \in [a,b],~$ then $$\int_a^b f(x)dx - \int_a^b g(x)dx = \int_a^b \left[ ~f(x) - g(x) ~\right] ~dx.$$ – user2661923 Apr 08 '23 at 04:55

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