Question

Исследовать на сходимость

Answer 1

$sumlimits _{n=1}^{+infty }, frac{5^{n}cdot (n+1)!}{(2n)!} \\Dalamber:; ; limlimits _{nto +infty } frac{a_{n+1}}{a{n}} =limlimits _{nto infty } left (frac{5^{n+1}cdot (n+2)!}{(2n+2)!}cdot frac{(2n)!}{5^{n}(n+1)!} right )=\\= limlimits _{nto infty } frac{5cdot (n+2)}{(2n+2)(2n+1)} = [frac{:n^2}{:n^2} ]=limlimits _{nto infty } frac{frac{5}{n}+frac{2}{n^2}}{4+frac{6}{n}+frac{2}{n^2}} =frac{0}{4}=0 textless 1; ; Rightarrow \\ryad; ; sxoditsya$

$P.S.; ; ; (2n+2)!=(2n)!cdot (2n+1)(2n+2)\\(n+2)!=(n+1)!cdot (n+2)$

Answer 2

Признак Даламбера.
$a_n=frac{5^n(n+1)!}{(2n)!}\a_{n+1}=frac{5^{n+1}(n+1+1)!}{(2(n+1))!}=frac{5cdot5^n(n+2)!}{(2n+2)!}\limlimits_{xtoinfty}frac{a_{n+1}}{a_n}=limlimits_{xtoinfty}left(frac{5cdot5^n(n+2)!}{(2n+2)!}:frac{5^n(n+1)!}{(2n)!}right)=\=limlimits_{xtoinfty}left(frac{5cdot5^n(n+2)!}{(2n+2)!}cdotfrac{(2n)!}{5^n(n+1)!}right)=limlimits_{xtoinfty}left(frac{5(n+2)}{(2n+1)(2n+2)}right)=$
$=limlimits_{xtoinfty}frac{5n+10}{4n^2+6n+2}=limlimits_{xtoinfty}frac{n(5+frac{10}n)}{n^2(4+frac6n+frac2{n^2})}=limlimits_{xtoinfty}frac{5}{4n}=0$

Ряд сходится.