How do you use the ratio test to test the convergence of the series $sum_(n=1)^oo (n!)/((2n+1)!)$ ?

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Answer 1 · 2018-05-19T06:53:17

The series converges

Let $u_n=(n!)/((2n+1)!)$

Then, the ratio test is

$|a_(n+1)/a_(n)|=|(((n+1)!)/((2(n+1)+1)!))/((n!)/((2n+1)!))|$

$=|((n+1)!)/(n!)((2n+1)!)/((2n+3)!)|$

$=|(n+1)/((2n+2)(2n+3))|$

$=|1/(2(2n+3))|$

$1/(2(2n+3))>0$ as $n in [1, +oo)$

Therefore,

$lim_(n->oo)|1/(2(2n+3))|=lim_(n->oo)1/(2(2n+3))$

$=0$

As the limit is $<1$ ,by the ratio test, the series converges

How do you use the ratio test to test the convergence of the series sum_(n=1)^oo (n!)/((2n+1)!)sum_(n=1)^oo (n!)/((2n+1)!)?