oVlad 21.04.2023 11:03 Let $n>4$ be a natural number. Prove that \[\sum_{k=2}^n\sqrt[k]{\frac{k}{k-1}}<n.\]
RagvaloD 21.04.2023 14:14 $(1+\frac{1}{k(k-1)})^k>1+\frac{1}{k-1} \to \sqrt[k]{\frac{k}{k-1}}<1+\frac{1}{k-1}-\frac{1}{k}$ So $\sum_{k=2}^n\sqrt[k]{\frac{k}{k-1}}<\sum_{k=2}^n (1+\frac{1}{k-1}-\frac{1}{k})=n-\frac{1}{n}$