Question

Исследовать ряды на сходимость:

Answer 1

$1)\; \; \; \sum \limits _{n=1}^{\infty } \frac{(n+1)!}{5^{n}}\\\\ \lim\limits _{n \to \infty}\frac{a_{n+1}}{a_{n}} = \lim\limits_{n \to \infty}\frac{(n+2)!}{5^{n+1}}: \frac{(n+1)!}{5^{n}} = \lim\limits _{n \to \infty} \frac{(n+1)!\, (n+2)}{5^{n}\cdot 5}\cdot \frac{5^{n}}{(n+1)!} =\\\\= \lim\limits _{n \to \infty} \frac{n+2}{5} =\infty \quad \Rightarrow \quad rasxoditsya$

$2)\; \; \; \sum \limits _{n=1}^{\infty }\Big (\frac{n}{3n-1}\Big )^{2n}\\\\ \lim\limits _{n \to \infty} \sqrt[n]{a_{n}} =\lim\limits _{n \to \infty} \sqrt[n]{ \Big (\frac{n}{3n-1}\Big )^{2n} }= \lim\limits_{n \to \infty}\Big (\frac{n}{3n-1}\Big )^2=(\frac{1}{3})^2= \frac{1}{9}\ \textless \ 1\; ,\\\\sxoditsya$

$3)\; \; \; \sum\limits _{n=1}^{\infty } \frac{(-1)^{n}\cdot n}{\sqrt{2n+1}} \\\\ \lim\limits _{n \to \infty} |a_n |= \lim\limits _{n \to \infty} \frac{n}{2n+1} = \frac{1}{2} \ne 0\; \; \Rightarrow \; \; rasxoditsya$