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[latex]\sum\limits _{n=1}^{+\infty }\, \frac{5^{n}\cdot (n+1)!}{(2n)!} \\\\Dalamber:\; \; \lim\limits _{n\to +\infty } \frac{a_{n+1}}{a{n}} =\lim\limits _{n\to \infty } \left (\frac{5^{n+1}\cdot (n+2)!}{(2n+2)!}\cdot \frac{(2n)!}{5^{n}(n+1)!} \right )=\\\\= \lim\limits _{n\to \infty } \frac{5\cdot (n+2)}{(2n+2)(2n+1)} = [\frac{:n^2}{:n^2} ]=\lim\limits _{n\to \infty } \frac{\frac{5}{n}+\frac{2}{n^2}}{4+\frac{6}{n}+\frac{2}{n^2}} =\frac{0}{4}=0\ \textless \ 1\; \; \Rightarrow \\\\ryad\; \; sxoditsya[/latex] [latex]P.S.\; \; \; (2n+2)!=(2n)!\cdot (2n+1)(2n+2)\\\\(n+2)!=(n+1)!\cdot (n+2)[/latex]

Признак Даламбера. [latex]a_n=\frac{5^n(n+1)!}{(2n)!}\\a_{n+1}=\frac{5^{n+1}(n+1+1)!}{(2(n+1))!}=\frac{5\cdot5^n(n+2)!}{(2n+2)!}\\\lim\limits_{x\to\infty}\frac{a_{n+1}}{a_n}=\lim\limits_{x\to\infty}\left(\frac{5\cdot5^n(n+2)!}{(2n+2)!}:\frac{5^n(n+1)!}{(2n)!}\right)=\\=\lim\limits_{x\to\infty}\left(\frac{5\cdot5^n(n+2)!}{(2n+2)!}\cdot\frac{(2n)!}{5^n(n+1)!}\right)=\lim\limits_{x\to\infty}\left(\frac{5(n+2)}{(2n+1)(2n+2)}\right)=[/latex] [latex]=\lim\limits_{x\to\infty}\frac{5n+10}{4n^2+6n+2}=\lim\limits_{x\to\infty}\frac{n(5+\frac{10}n)}{n^2(4+\frac6n+\frac2{n^2})}=\lim\limits_{x\to\infty}\frac{5}{4n}=0[/latex] Ряд сходится.