Answer:
Given series is divergent
Explanation:
Step(i):-
By using Ratio test
![\lim_(n \to \infty) |(a_(n+1) )/(a_(n) ) | = l](https://img.qammunity.org/2021/formulas/mathematics/college/hlnixrg5hqqe9ums3mcnip2aqp151m96se.png)
a)
'l' is finite then the given ∑aₙ is convergent
b)
Here 'l' is infinite then the ∑aₙ is divergent
Step(ii):-
Given aₙ =
![(n!)/(n)](https://img.qammunity.org/2021/formulas/mathematics/college/hf4eql8o8jj8kr5efeyun9okz9s1u3t4n6.png)
![a_(n+1) = ((n+1)!)/(n+1)](https://img.qammunity.org/2021/formulas/mathematics/college/bv1uvkwstw2fzvhajfhuforq8ulqtksh7s.png)
![\lim_(n \to \infty) |(a_(n+1) )/(a_(n) ) | = \lim_(n \to \infty) |(((n+1)!)/(n+1) )/((n!)/(n) ) |](https://img.qammunity.org/2021/formulas/mathematics/college/eazzh9mclsq1nf0gdpbzeg7iucw6xe8rp5.png)
we know that n ! = n (n-1) (n-2) ......3.2.1
and also (n+1) ! = (n+1)n!
![\lim_(n \to \infty) |(a_(n+1) )/(a_(n) ) | = \lim_(n \to \infty) |((n+1)n!)/(n+1) )/((n!)/(n) ) |](https://img.qammunity.org/2021/formulas/mathematics/college/kpi9dx5affznzx4ebqaihiy5hpjqc2btv2.png)
![\lim_(n \to \infty) |(a_(n+1) )/(a_(n) ) | = \lim_(n \to \infty) n](https://img.qammunity.org/2021/formulas/mathematics/college/aujf0fh0821wvbcwwtp430dqa7szq8z11e.png)
= ∞
Given sum of the series is divergent