Question

\( \operatorname{Lim}_{n \rightarrow \infty} \frac{\left(1^{2022}+2^{2022}+3^{2022}+\ldots n^{2022}\right) n}{1^{2023}+2^{2023}+\ldots n^{2023}} \) is equal to \( \frac{a}{b} \) where \( a \) and \( b \) are coprime then \( a+b= \) (A) 4045 (B) 4046 (C) 4047 (D) None of these

Answer 1

\(\lim\limits_{n\to\infty} \frac{1^{2022} + 2^{2022} + 3^{2022}+....+n^{2022}}{1^{2023} + 2^{2023}+ 3^{2023} + ....+n^{2023}}\)

\(= \lim\limits_{n\to\infty} \cfrac{\sum\limits^n_{r = 1}r^{2022}}{\sum\limits^n_{r = 1}r^{2023}}\)

\(= \lim\limits_{n\to\infty} \cfrac{\frac1n\sum\limits^n_{r = 1}r^{2022}}{\frac1n\sum\limits^n_{r = 1}r^{2023}}\)

\(= \cfrac{\int\limits^1_0x^{2022}dx}{\int\limits_{0}^1 x^{2023}dx}\)

\(= \cfrac{\left[\frac{x^{2023}}{2023}\right]^1_0}{\left[\frac{x^{2024}}{2024}\right]^1_0}\)

\(= \frac{2024}{2023}\)

\(\therefore a + b = 2024 + 2023 = 4047\).