№2125
0
$$\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{3}}}} } \right)}^3} \cdot {{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{4}}}} } \right)}^4}}}{{{{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{2}}}} } \right)}^2} \cdot {{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{5}}}} } \right)}^5}}}$$
комментарии

Your solution