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# How do you apply the ratio test to determine if $\sum{\dfrac{{{3}^{n}}{{(n!)}^{2}}}{(2n)!}}$ from $[1,\infty )$ is convergent or divergent?

Last updated date: 22nd Jul 2024
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Hint: The given expression is written in the formula for the ratio test. The determination of the expression whether it is convergent, divergent or inconclusive is conditional and is based on the value of the limit we will get on solving the ratio test equation.

According to the question, we have to check whether the expression is convergent or divergent. Before moving on with the solution, we will first try to understand the question.
The given question can also be written as:
$\sum\limits_{n=1}^{\infty }{\dfrac{{{3}^{n}}{{(n!)}^{2}}}{(2n)!}}$
So, the question says how will the expression respond when the value of $n$ increases, that is when the value of $n$ approaches infinity. For this, we will be using ratio test, which can be stated as follows:
For an expression in the form $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ where the value of $n$ approaches infinity then limiting the value of $n$ as $n \to \infty$ using the formula : $\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|=L$
And if $L<1$, then the series converges
for $L>1$, the series diverges
and when $L=1$, then it is inconclusive or cannot say how it will respond.

We can also see factorials in our question. So, factorials are expressed with a (!) sign.
For example - $a!=a\times (a-1)\times (a-2)...$
Now, we are ready to start solving the question.
The expression we have is $\sum\limits_{n=1}^{\infty }{\dfrac{{{3}^{n}}{{(n!)}^{2}}}{(2n)!}}$, writing this expression in the formula for ratio test $\left( \displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|=L \right)$, we have
First, we will write the value for ${{a}_{n+1}}$ and ${{a}_{n}}$,
${{a}_{n+1}}=\dfrac{{{3}^{n+1}}{{((n+1)!)}^{2}}}{(2(n+1))!}$ and ${{a}_{n}}=\dfrac{{{3}^{n}}{{(n!)}^{2}}}{(2n)!}$
Substituting in the expression of ratio test, we have
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{3}^{n+1}}{{((n+1)!)}^{2}}}{(2(n+1))!}}{\dfrac{{{3}^{n}}{{(n!)}^{2}}}{(2n)!}} \right|=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{3}^{n+1}}{{((n+1)!)}^{2}}}{(2(n+1))!}\times \dfrac{(2n)!}{{{3}^{n}}{{(n!)}^{2}}} \right|$
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{3}^{n}}.3{{((n+1)!)}^{2}}}{(2n+2))!}\times \dfrac{(2n)!}{{{3}^{n}}{{(n!)}^{2}}} \right|$
Now we open up the factorials to cancel the similar terms,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{3{{((n+1)(n!))}^{2}}}{(2n+2)(2n+1)(2n)!}\times \dfrac{(2n)!}{{{(n!)}^{2}}} \right|$
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{3{{(n+1)}^{2}}{{(n!)}^{2}}}{(2n+2)(2n+1)}\times \dfrac{1}{{{(n!)}^{2}}} \right|$
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{3{{(n+1)}^{2}}}{(2(n+1))(2n+1)}\times \dfrac{1}{1} \right|$
$\Rightarrow \dfrac{3}{2}\displaystyle \lim_{n \to \infty }\left| \dfrac{{{(n+1)}^{2}}}{(n+1)(2n+1)} \right|=\dfrac{3}{2}\displaystyle \lim_{n \to \infty }\left| \dfrac{(n+1)}{(2n+1)} \right|$
Multiplying and dividing by $1/n$, we get
$\Rightarrow \dfrac{3}{2}\displaystyle \lim_{n \to \infty }\left| \dfrac{(n+1)}{(2n+1)}\times \dfrac{1/n}{1/n} \right|=\dfrac{3}{2}\displaystyle \lim_{n \to \infty }\left| \dfrac{(1+(1/n))}{(2+(1/n))} \right|$
$\Rightarrow \dfrac{3}{2}\displaystyle \lim_{n \to \infty }\left| \dfrac{(1+(1/n))}{(2+(1/n))} \right|=\dfrac{3}{2}\left| \dfrac{(1+(0)}{(2+(0))} \right|$
$\Rightarrow \dfrac{3}{2}\left( \dfrac{1}{2} \right)=\dfrac{3}{4}$
which is less than 1 as $\left( \dfrac{3}{4}<1 \right)$.
According to the ratio test, if the value of the limit is less than 1 then, the expression is convergent.
Therefore, the expression is convergent.

Note: Apply the factorial expansion correctly else extra terms will add to the complexity of the expression. Expand the factorial in such a way that similar factorials get cancelled. Recall the conditions in the ratio test which determine whether the expression is convergent or divergent or none.