1. Masalani bayon qilamiz: \(\sum_{n=1}^{\infty} \frac{n!}{n^{n-1}}\) ketma-ketligi yig'indisi mavjudligini tekshirish va yaqinlashishni aniqlash. 2. Ketma-ketlikdagi har bir hadni ko'rib chiqamiz: \(a_n = \frac{n!}{n^{n-1}}\). 3. Yig'indining yaqinlashishini tekshirish uchun nisbiy testdan foydalanamiz. Nisbiy test formulasi: $$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$ 4. Hisoblaymiz: $$\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^n}}{\frac{n!}{n^{n-1}}} = \frac{(n+1)!}{(n+1)^n} \cdot \frac{n^{n-1}}{n!} = \frac{(n+1) \cdot n^{n-1}}{(n+1)^n} = \frac{n^{n-1}}{(n+1)^{n-1}} = \left(\frac{n}{n+1}\right)^{n-1}$$ 5. Limitni topamiz: $$L = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n-1} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n-1} = e^{-1} = \frac{1}{e}$$ 6. Chunki \(L = \frac{1}{e} < 1\), nisbiy testga ko'ra ketma-ketlik yig'indisi yaqinlashadi. 7. Natija: \(\sum_{n=1}^{\infty} \frac{n!}{n^{n-1}}\) ketma-ketligi yaqinlashadi. Bu ketma-ketlikning aniq yig'indisi oddiy ifoda bilan ifodalanmaydi, lekin yaqinlashishi tasdiqlandi.