Question

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

Answer 1

Answer:

    $\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑$(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let   $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

   $U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

 $U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

    $\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

                       $= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

                    = $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

                       $=\frac{1}{infinite }$

                     =0

    $\lim_{n \to \infty} U_n =0$

∴ Given series is converges