Is $(2n)!$ the same as $2(n!)$?
I am trying to determine the convergence of a series $$\sum_{n=17}^{\infty} \frac{(n!)}{(2n)!}.$$ Using the ratio test, I have simplified $a_{n+1}/a_n$ to $$\frac{(2n!)}{2(n)!}.$$ If $(2n)!$ is the same as $2(n!)$, I can strip out the factorial to factor the ratio to $1$. I do acknowledge that they are most likely not the same as $2(n!)$ is multiplying the result of $n!$, but I am not sure how to proceed.
$\endgroup$ 66 Answers
$\begingroup$$${a_{n+1}\over a_n}={(n+1)!\over n!}\cdot {(2n)!\over(2n+2)!}={n+1\over(2n+2)(2n+1)}={1\over4n+2}$$
$\endgroup$ $\begingroup$We have$$\sum _{n=m}^\infty \frac{n!}{n!(n+1)(n+2)\cdots 2n} \leq \sum _{n=m}^\infty \frac{1}{n^2} < \infty .$$Therefore the series $\sum _{n=m}^\infty \frac{n!}{(2n)!}$ converges.
$\endgroup$ $\begingroup$In general, $(2n)!$ is enormously larger than $n!$. You may notice that$$ \sum_{n\geq 0}\frac{n!}{(2n)!} = \int_{0}^{+\infty}e^{-x}\sum_{n\geq 0}\frac{x^n}{(2n)!}\,dx=\int_{0}^{+\infty}e^{-x}\cosh(\sqrt{x})\,dx=\int_{0}^{+\infty}x(e^x+e^{-x})e^{-x^2}\,dx $$so$$ \sum_{n\geq 0}\frac{n!}{(2n)!} = 1+\frac{\sqrt{\pi}e^{1/4}}{2}\text{Erf}\left(\frac{1}{2}\right)<1+\frac{\sqrt{\pi}}{2}e^{1/4} $$by completing the square. As an alternative, $\binom{2n}{n}\geq \frac{4^n}{n+1}$ for any $n\geq 1$ is granted by the Cauchy-Schwarz inequality, so$$ \sum_{n\geq 0}\frac{n!}{(2n)!} = 1+\sum_{n\geq 1}\frac{1}{n!\binom{2n}{n}}\leq1+\sum_{n\geq 1}\frac{n+1}{n! 4^n}=\frac{5}{4}e^{1/4}. $$
$\endgroup$ $\begingroup$In math it is almost never the case that you can move something out of parenthis.
$(2x)^2 \ne 2(x^2)$ and likewise $(2n)!\ne 2(n!)$
And this can be easily varified.
$(2n)! = 1*2*3*.......*2n$
Whereas $2(n!) = 2\times (1*2*3*.......*n)$
And very simple case:
$(2\cdot 3)! = 6! = 1*2*3*4*5*6 = 720$
And $2(3!) = 2\times (1*2*3) = 2*6 =12$.
Indeed you always get $(2n)! = (1*2*3*.....*n)*([n+1]*[n+2]*...... *2n)$ while
$2(n!) = (1*2*3*......*n)*2$
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Okay, enough beating a dead horse.
You can compare $\frac {a_{n+1}}{a_n} = \frac {\frac {(n+1)!}{(2(n+1))!}}{\frac {n!}{(2n)!}}=$
$\frac {(n+1)!}{(2n+2)!}\cdot \frac {(2n)!}{n!}=$
$\frac {1*2*.......*n*(n+1)}{1*2*......*2n*(2n+1)*(2n+2)}\cdot \frac {1*2*3*....*2n}{1*2*....*n}=$
$\frac {n+1}{(2n+1)(2n+2)}=$
$\frac {n+1}{(2n+1)\cdot2\cdot(n+1)}= \frac 1{2(2n+1)}$.
Not sure how you did your simplification.
$\endgroup$ 1 $\begingroup$No they are not the same, try with $n=3$
$$\frac{(2n)!}{2(n)!}=\frac{(6!)}{2(3)!}=6\cdot 5 \cdot 2$$
which indicates that $(2n)!$ is much larger than $2(n)!$.
We don't need ratio test in this case indeed by direct comparison we see that
$$\sum_{n=1}^{\infty} \frac{(n!)}{(2n)!}\le \sum_{n=1}^{\infty} \frac{1}{2n(2n-1)}\le \sum_{n=1}^{\infty} \frac{1}{2n^2}=\frac {\pi^2} {12}$$
$\endgroup$ 0 $\begingroup$Ratio test is fine.$$\frac {a_{n+1}}{a_n}=\frac {(n+1)!(2n)!}{(2n+2)(2n+1)(2n)!(n!)} =$$
$$ \frac {1}{2(2n+1)}\to 0 $$
Thus the series converge.
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