证明: 当 $p\geq1$ 时, $$\bex \vsm{n}\frac{1}{(n+1)\sqrt[p]{n}}<p. \eex$$ (国外赛题)

证明: 通项 $$\beex \bea a_n&=\frac{1}{(n+1)\sqrt[p]{n}} =\frac{n}{\sqrt[p]{n}}\cdot \frac{1}{n(n+1)} =n^\frac{p-1}{p}\sex{\frac{1}{n}-\frac{1}{n+1}}\\ &=n^\frac{p-1}{p} \sez{\sex{\frac{1}{\sqrt[p]{n}}}^p -\sex{\frac{1}{\sqrt[p]{n+1}}}^p}\\ &=n^\frac{p-1}{p} \cdot p\sex{\frac{1}{\sqrt[p]{n+\tt}}}^{p-1} \sex{ \frac{1}{\sqrt[p]{n}}-\frac{1}{\sqrt[p]{n+1}}}\quad\sex{0<\tt<1}\\ &<p\sez{\frac{1}{\sqrt[p]{n}}-\frac{1}{\sqrt[p]{n+1}}} \quad\sex{\frac{n^\frac{p-1}{p}}{\sex{\sqrt[p]{n+\tt}}^{p-1}}<1}, \eea \eeex$$ 而前 $n$ 项和 $$\bex S_n=a_1+\sum_{k=2}^n a_k <a_1+\sum_{k=2}^n p\sez{\frac{1}{\sqrt[p]{n}}-\frac{1}{\sqrt[p]{n+1}}} =a_1+p\sex{\frac{1}{\sqrt[p]{2}}-\frac{1}{\sqrt[p]{n+1}}}, \eex$$ $$\bex \vlm{n}S_n\leq a_1+\frac{p}{\sqrt[p]{2}} <p\sex{1-\frac{1}{\sqrt[p]{2}}}+\frac{p}{\sqrt[p]{2}}=p. \eex$$