Answer:
[tex]\lim_{n \to \infty} U_n =0[/tex]
Given series is convergence by using Leibnitz's rule
Step-by-step explanation:
Step(i):-
Given series is an alternating series
∑[tex](-1)^{n} \frac{n^{2} }{n^{3}+3 }[/tex]
Let [tex]U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }[/tex]
By using Leibnitz's rule
[tex]U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }[/tex]
[tex]U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}[/tex]
Uₙ-Uₙ₋₁ <0
Step(ii):-
[tex]\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }[/tex]
[tex]= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }[/tex]
= [tex]= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }[/tex]
[tex]=\frac{1}{infinite }[/tex]
=0
[tex]\lim_{n \to \infty} U_n =0[/tex]
∴ Given series is converges
