How do you use the ratio test to test the convergence of the series #∑ [n(n!)^2]/(2n+1)!# from n=1 to infinity?

Question

Scarlett Allison · Accepted Answer

The series:

#sum_(n=1)^oo (n(n!)^2)/((2n+1)!)#

is convergent.

We have to analyse: #lim_(n->oo) abs (a_(n+1)/a_n)# so we first calculate the ratio: #abs (a_(n+1)/a_n) = frac ( ( ( n+1) ((n+1)!)^2)/((2(n+1)+1)!)) ( (n(n!)^2)/((2n+1)!)) = (n+1)/n ((n+1)!)^2/(n!)^2 ((2n+1)!)/((2n+3)!)# #abs (a_(n+1)/a_n) = (n+1)^3/(n(2n+3)(2n+2)) = (n^3+3n^2+3n+1)/(4n^3+10n^2+6n)# and we can see that: #lim_(n->oo) abs (a_(n+1)/a_n) =1/4 < 1# which means the series is convergent.

Charles Arocha · Answer

undefined To test the convergence of the series $\sum \frac{n(n!)^2}{(2n+1)!}$ from $n=1$ to infinity using the ratio test, we compute the limit:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

where $a_n$ represents the general term of the series.

In this case, $a_n = \frac{n(n!)^2}{(2n+1)!}$. So, we need to find:

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

Simplify this expression:

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

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

$$ = \lim_{n \to \infty} \left| \frac{n^3 + 3n^2 + 3n + 1}{4n^2 + 10n + 6} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{n^3}{4n^2} \right| $$

$$ = \lim_{n \to \infty} \frac{n}{4} $$

$$ = \infty $$

Since the limit is greater than 1, according to the ratio test, the series diverges.