Question
Suppose that $E|X|^r<\infty$ for some $r>0$. Then $x^rP(|X|\geq x)\to 0$ as $x\to \infty$. Conversely if $x^rP(|X|\geq x)\to 0$ as $x\to \infty$, then $E|X|^s<\infty$ for $0\leq s<r$.
My attempt
By Markov's inequality if $E|X|^r<\infty$ for some $r>0$, then $$ 0\leq x^{r-1}P(|X|\geq x)\leq x^{r-1}\times\frac{E|X|^r}{x^r}\to0 $$ as $x\to \infty$. I am also aware that $$ E|X|^r=\int_{0}^\infty rx^{r-1} P(|X|\geq x)\, dx $$ but I have been unable to prove the stronger result that $x^rP(|X|\geq x)\to0$ as $x\to \infty$. Any help is appreciated.
For the first part, what you did proves that $x^{r-1}P(|X|\geq x)$, but we want more. This can be done in the following way: $x^r\mathbf 1\{|X|\geq x\}\leq |X|^r\mathbf 1\{|X|\geq x\} $. Then take the expectations and use the monotone convergence theorem.
For the second part, the formula you mention can be used and combined with the bound $x^{s-1}P(|X|\geq x)\leq \min\left\{x^{s-1},x^{s-r-1}\right\}$ .