How do you apply the ratio test to determine if #Sigma (3^n(n!)^2)/((2n)!)# from #n=[1,oo)# is convergent to divergent?

Question

Christopher Agresta · Accepted Answer

Prove that: #lim_(n->oo) abs (a_(n+1)/a_n) <1#

The ratio test states that a necessary condition for the series:

#sum_(n=1)^oo a_n#

to converge is that

#L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1#

If #L<1# the condition is also sufficient, while for #L=1# the test is inconclusive.

Calculate the expression of the ratio for this series:

#abs (a_(n+1)/a_n) = ( (3^(n+1) ((n+1)!)^2) / (2(n+1)!)) / ( (3^n (n!)^2)/ ((2n)!)) = 3^(n+1)/3^n ((n+1)!)^2/((n!)^2) ((2n)!)/ ((2(n+1))!)#

#abs (a_(n+1)/a_n) = 3 (n+1)^2/((2n+2)(2n+1))=3/2 (n+1)/(2n+1)#

Passing to the limit:

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)3/2 (n+1)/(2n+1) = 3/4 < 1#

so the series is convergent.

Emilia Aguilar · Answer

undefined To apply the ratio test to determine the convergence or divergence of the series $ \sum_{n=1}^\infty \frac{3^n(n!)^2}{(2n)!} $, calculate the limit as $ n $ approaches infinity of the absolute value of the ratio of the $ (n+1) $th term to the $ n $th term. If the limit is less than 1, the series is convergent; if it's greater than 1, the series is divergent; if it equals 1, the test is inconclusive.

Using the ratio test:

$$
\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\frac{3^{n+1}((n+1)!)^2}{(2(n+1))!}}{\frac{3^n(n!)^2}{(2n)!}} \right|
$$

After simplification:

$$
\lim_{{n \to \infty}} \left| \frac{3^{n+1}(n+1)^2(n+1)!^2(2n)!}{(2(n+1))!3^nn!^22n!} \right|
$$

This simplifies to:

$$
\lim_{{n \to \infty}} \left| \frac{3(n+1)^2(n+1)!^2}{(2n+2)(2n+1)(2n+2)(2n+1)} \right|
$$

This further simplifies to:

$$
\lim_{{n \to \infty}} \left| \frac{3(n+1)(n+1)!^2}{(2n+2)^2(2n+1)^2} \right|
$$

Now, as $ n $ approaches infinity, the limit of this expression can be evaluated. If the limit is less than 1, the series converges; if it's greater than 1, the series diverges.

After calculation, if the limit is less than 1, then the series $ \sum_{n=1}^\infty \frac{3^n(n!)^2}{(2n)!} $ is convergent. If it's greater than 1, the series is divergent. If it equals 1, the test is inconclusive.