# How do you test for convergence for 1/((2n+1)!) ?

Jul 30, 2015

In the case you meant "test the convergence of the series : sum_(n=1)^(oo)1/((2n+1)!)"

the Answer is : it $\textcolor{b l u e}{\text{converges}}$

#### Explanation:

To find out, we can use the ratio test.
That is, if $\text{U"_"n}$ is the ${n}^{\text{th}}$ term of this series

Then if, we show that ${\lim}_{n \rightarrow + \infty} \left\mid {\text{U"_("n"+1)/"U}}_{n} \right\mid < 1$
it means that the series converges

On the other if lim_(nrarr+oo)abs(("U"_("n"+1))/"U"_n)>1
it means that the series diverges

In our case

"U"_n=1/((2n+1)!)

$\text{ }$ and

"U"_("n"+1)=1/([2(n+1)+1]!)=1/([2n+3]!)

Hence, "U"_("n"+1)/"U"_n=1/((2n+3)!)÷1/((2n+1)!)=((2n+1)!)/((2n+3)!)

$\text{Notice that} :$
(2n+3)! =(2n+3)xx(2n+2)xx(2n+1)!

Just like : 10! =10xx9xx8!
We subtract $1$ each time to get the next

So we have,
"U"_("n"+1)/"U"_n=((2n+1)!)/((2n+3)(2n+2)(2n+1)!)=1/((2n+3)(2n+2))

Next we test,

${\lim}_{n \rightarrow + \infty} \left\mid {\text{U"_("n"+1)/"U}}_{n} \right\mid$

$= {\lim}_{n \rightarrow + \infty} \left\mid \frac{1}{\left(2 n + 3\right) \left(2 n + 2\right)} \right\mid = {\lim}_{n \rightarrow + \infty} \frac{1}{\left(4 {n}^{2} + 10 n + 6\right)} = \frac{1}{+ \infty} = 0 \text{ }$ and $0$ is less than $1$

Hence, it's quite safe to conclude that the series color(blue)"converges" !