Question

Решите метожом даламбера

Answer 1

$\sum \limits _{n=1}^{\infty }\, \frac{n^{n}}{(2n)!}\\\\\\\lim\limits _{n \to \infty}\frac{(n+1)^{n+1}}{(2n+2)!}:\frac{n^{n}}{(2n)!}=\lim\limits _{n \to \infty}\frac{(n+1)^{n}\cdot (n+1)}{(2n)!\cdot (2n+1)\cdot (2n+2)}\cdot \frac{(2n)!}{n^{n}}=\\\\=\lim\limits _{n \to \infty}\Big (\Big (\frac{n+1}{n}\Big )^{n}\cdot \frac{n+1}{(2n+1)(2n+2)}\Big )=\lim\limits _{n \to \infty}\Big (\Big (1+\frac{1}{n}\Big )^{n}\cdot \frac{n}{4n}\Big )=\frac{1}{4}\cdot e\approx 0,68<1\\\\\\\Rightarrow \; \; ryad\; sxoditsya$