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

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

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Answer 1 · 2018-05-19T06:53:17

The series converges

Explanation:

Let $u_{n} = \frac{n!}{(2 n + 1)!}$ $u_n=(n!)/((2n+1)!)$

Then, the ratio test is

$∣ ∣ ∣ \frac{a_{n + 1}}{a_{n}} ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \frac{\frac{(n + 1)!}{(2 (n + 1) + 1)!}}{\frac{n!}{(2 n + 1)!}} ∣ ∣ ∣ ∣ ∣$ $|a_(n+1)/a_(n)|=|(((n+1)!)/((2(n+1)+1)!))/((n!)/((2n+1)!))|$

$= ∣ ∣ ∣ \frac{(n + 1)!}{n!} \frac{(2 n + 1)!}{(2 n + 3)!} ∣ ∣ ∣$ $=|((n+1)!)/(n!)((2n+1)!)/((2n+3)!)|$

$= ∣ ∣ ∣ \frac{n + 1}{(2 n + 2) (2 n + 3)} ∣ ∣ ∣$ $=|(n+1)/((2n+2)(2n+3))|$

$= ∣ ∣ ∣ \frac{1}{2 (2 n + 3)} ∣ ∣ ∣$ $=|1/(2(2n+3))|$

$\frac{1}{2 (2 n + 3)} > 0$ $1/(2(2n+3))>0$ as $n \in [1, + \infty)$ $n in [1, +oo)$

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

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

1 Answer

Explanation:

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