Cho dãy số $(u_n)$ biết $u_n = \frac{1}{(n+1)\sqrt{n} + n\sqrt{n+1}}$, $n \in \mathbb{N}^*$. Với mỗi số nguyên dương $n$, đặt $S_n = u_1 + u_2 + \dots + u_n$. Tìm $n$ để $S_n + \frac{1}{\sqrt{2024}} = 1$.

Lời giải Ta có: $$ u_n = \frac{1}{(n+1)\sqrt{n} + n\sqrt{n+1}} = \frac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} + \sqrt{n})} $$ $$ = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n} \cdot \sqrt{n+1} \cdot (\sqrt{n+1} + \sqrt{n}) \cdot (\sqrt{n+1} - \sqrt{n})} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n} \cdot \sqrt{n+1}} $$ $$ = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}, \quad \forall n \in \mathbb{N}^* $$ Suy ra: $$ S_n = u_1 + u_2 + \dots + u_n = \Big(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}\Big) + \Big(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\Big) + \dots + \Big(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\Big) = 1 - \frac{1}{\sqrt{n+1}} $$ Ta có: $$ S_n + \frac{1}{\sqrt{2024}} = 1 \Leftrightarrow 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{2024}} = 1 $$ $$ \Leftrightarrow \frac{1}{\sqrt{n+1}} = \frac{1}{\sqrt{2024}} \Leftrightarrow n = 2023 $$