Problem 1. Let $a_{0}, a_{1}, \ldots, a_{k}(k \geq 1)$ be real numbers such that $a_{0}+a_{1}+\ldots+a_{k}=0$. Calculate $\lim _{n \rightarrow \infty}\left(a_{0} \sqrt[3]{n}+a_{1} \sqrt[3]{n+1}+\ldots+a_{k} \sqrt[3]{n+k}\right)$.
Solution. Substitute $a_{0}=-a_{1}-a_{2}-\ldots-a_{k}$ and thus $a_{0} \sqrt[3]{n}+a_{1} \sqrt[3]{n+1}+\ldots+a_{k} \sqrt[3]{n+k}=$ $a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+\ldots+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})$.

However, for any $p=1,2, \ldots, k, \sqrt[3]{n+p}-\sqrt[3]{n}=\frac{n+p-n}{\sqrt[3]{(n+p)^{2}}+\sqrt[3]{(n+p) n}+\sqrt[3]{n^{2}}}=$ $=\frac{p}{\sqrt[3]{(n+p)^{2}}+\sqrt[3]{(n+p) n}+\sqrt[3]{n^{2}}} \longrightarrow 0$ as $n \rightarrow \infty$.

Therefore, the sought limit is 0.