How do you test for convergence for $\frac{1}{(2 n + 1)!}$ $1/((2n+1)!)$ ?

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Answer 1 · 2015-07-30T19:28:28

In the case you meant "test the convergence of the series : $\infty \sum n = 1 \frac{1}{(2 n + 1)!}$ $sum_(n=1)^(oo)1/((2n+1)!)$ "

the Answer is : it $converges$ $color(blue)"converges"$

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

Then if, we show that $lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)<1$
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

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

$" "$ and

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

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

$"Notice that" :$
$(2n+3)! =(2n+3)xx(2n+2)xx(2n+1)!$

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,

$lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)$

$=lim_(nrarr+oo)abs(1/((2n+3)(2n+2)))=lim_(nrarr+oo)1/((4n^2+10n+6))=1/(+oo)=0" "$ and $0$ is less than $1$

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

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