Question

Test the series for convergence or divergence (using ratio test)​

Answer 1

`\lim_(n \to \infty) U_n =0`

Given series is convergence by using Leibnitz's rule

Explanation:

Step(i):-

Given series is an alternating series

∑
`(-1)^(n) (n^(2) )/(n^(3)+3 )`

Let
`U_(n) = (-1)^(n) (n^(2) )/(n^(3)+3 )`

By using Leibnitz's rule

`U_(n) - U_(n-1) = (n^(2) )/(n^(3) +3) - ((n-1)^(2) )/((n-1)^(3)+3 )`

`U_(n) - U_(n-1) = (n^(2)(n-1)^(3) +3)-(n-1)^(2) (n^(3) +3) )/((n^(3) +3)(n-1)^(3) +3))`

Uₙ-Uₙ₋₁ <0

Step(ii):-

`\lim_(n \to \infty) U_n = \lim_(n \to \infty)(n^(2) )/(n^(3)+3 )`

`= \lim_(n \to \infty)(n^(2) )/(n^(3)(1+(3)/(n^(3) ) ) )`

=
`= \lim_(n \to \infty)(1 )/(n(1+(3)/(n^(3) ) ) )`

`=(1)/(infinite )`

=0

`\lim_(n \to \infty) U_n =0`

∴ Given series is converges