How do you use the ratio test to test the convergence of the series $\sum{\dfrac{{{\left( n! \right)}^{2}}}{kn}!}$ from $n=1$ to infinity?
Answer
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Hint: Convergence, in mathematics, is a property of approaching a limit when the variable in a function increases or decreases. It is exhibited by some of the infinite functions and series. The ratio test states that the necessary condition for the series $\sum\nolimits_{n=1}^{\infty }{{{x}_{n}}}$ to converge is given by $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|<1$.
Complete step by step solution:
We are given a series from $n=1$ to infinity and need to test the convergence of the series. We will be using the ratio test to test the convergence of the given series.
The ratio test states that the condition for the series $\sum\nolimits_{n=1}^{\infty }{{{x}_{n}}}$ to converge is given by
$L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|<1$
The given series is convergent if $L<1$
In our question, the series given is $\sum{\dfrac{{{\left( n! \right)}^{2}}}{kn}!}$ from $n=1$ to infinity.
Here,
${{x}_{n}}=\sum{\dfrac{{{\left( n! \right)}^{2}}}{kn}!}$ ;
The value of ${{x}_{n+1}}$ is obtained by replacing the value of $n$ with $n+1$
Following the same, we get,
$\Rightarrow {{x}_{n+1}}=\left( \dfrac{{{\left( n+1 \right)}^{2}}!}{\left( k\left( n+1 \right) \right)!} \right)$
Firstly, we need to solve for $\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|$
$\Rightarrow \left| \dfrac{{{x}_{n+1}}}{\dfrac{{{\left( n! \right)}^{2}}}{\left( kn \right)!}} \right|$
Substituting the value of ${{x}_{n+1}}$from above, we get,
$\Rightarrow \left| \dfrac{\left( \dfrac{{{\left( n+1 \right)}^{2}}!}{\left( k\left( n+1 \right) \right)!} \right)}{\dfrac{{{\left( n! \right)}^{2}}}{\left( kn \right)!}} \right|$
Let us evaluate further.
$\Rightarrow \dfrac{{{\left( \left( n+1 \right)! \right)}^{2}}}{{{\left( n! \right)}^{2}}}\times \dfrac{\left( kn \right)!}{\left( kn+k \right)!}$
Simplifying the above equation, we get,
$\Rightarrow \dfrac{{{\left( n+1 \right)}^{2}}}{\left( kn+k \right)\left( kn+k-1 \right)......\left( kn+1 \right)}$
$\therefore \left| \dfrac{{{x}_{n}}}{{{x}_{n+1}}} \right|=\dfrac{{{\left( n+1 \right)}^{2}}}{\left( kn+k \right)\left( kn+k-1 \right)......\left( kn+1 \right)}$
Now, we have to find out the value of the limit for different values of k.
Following the same, for $k=1$, the value of the limit is
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{n+1}$
Cancelling the common factors on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\left( n+1 \right)$
Evaluating the limit,
$\therefore \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=+\infty$ for $k=1$
For $k=2$,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{\left( n+1 \right)\left( n+2 \right)}$
Cancelling the common factors on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{{}}}}{\left( n+2 \right)}$
Dividing the numerator and denominator with $n$ on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( 1+\dfrac{1}{n} \right)}^{{}}}}{\left( 1+\dfrac{2}{n} \right)}$
As $n \to \infty$, $\dfrac{1}{n}\to \infty$. Following the same,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=1$
The value of the limit is equal to one and the result is inconclusive. So, we have to further evaluate the series $\sum\nolimits_{n=1}^{\infty }{\dfrac{{{\left( n! \right)}^{2}}}{\left( 2n \right)!}}$
Expanding the factorial terms, we get,
$\Rightarrow \sum\nolimits_{n=1}^{\infty }{\dfrac{{{\left( n! \right)}^{2}}}{\left( 2n \right)!}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}$
From series and sequences,
$\Rightarrow \dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}=\prod\limits_{p=1}^{n}{\dfrac{p}{n+p}}$
From the above $p\le n$ ,
$\Rightarrow \dfrac{p}{n+p}\le \dfrac{p}{p+p}\le \dfrac{p}{2p}\le \dfrac{1}{2}$
Substituting the value on the right-hand side, we get,
$\Rightarrow \dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}\le {{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow \sum\nolimits_{n=1}^{\infty }{{{\left( \dfrac{1}{2} \right)}^{n}}=1}$
The above series is convergent, so the given series is also convergent.
Now,
For the value of $k\ge 3$ ,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{\left( n+1 \right)\left( n+2 \right)....\left( n+k \right)}$
$\therefore \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=0$
$\therefore$The given series is convergent for the value $k\ge 2$
Note: One must remember that the multiplication happens to a given number down to the number one and not zero. Factorial of a number in mathematics is the product of all the positive numbers less than or equal to a number.
Complete step by step solution:
We are given a series from $n=1$ to infinity and need to test the convergence of the series. We will be using the ratio test to test the convergence of the given series.
The ratio test states that the condition for the series $\sum\nolimits_{n=1}^{\infty }{{{x}_{n}}}$ to converge is given by
$L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|<1$
The given series is convergent if $L<1$
In our question, the series given is $\sum{\dfrac{{{\left( n! \right)}^{2}}}{kn}!}$ from $n=1$ to infinity.
Here,
${{x}_{n}}=\sum{\dfrac{{{\left( n! \right)}^{2}}}{kn}!}$ ;
The value of ${{x}_{n+1}}$ is obtained by replacing the value of $n$ with $n+1$
Following the same, we get,
$\Rightarrow {{x}_{n+1}}=\left( \dfrac{{{\left( n+1 \right)}^{2}}!}{\left( k\left( n+1 \right) \right)!} \right)$
Firstly, we need to solve for $\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|$
$\Rightarrow \left| \dfrac{{{x}_{n+1}}}{\dfrac{{{\left( n! \right)}^{2}}}{\left( kn \right)!}} \right|$
Substituting the value of ${{x}_{n+1}}$from above, we get,
$\Rightarrow \left| \dfrac{\left( \dfrac{{{\left( n+1 \right)}^{2}}!}{\left( k\left( n+1 \right) \right)!} \right)}{\dfrac{{{\left( n! \right)}^{2}}}{\left( kn \right)!}} \right|$
Let us evaluate further.
$\Rightarrow \dfrac{{{\left( \left( n+1 \right)! \right)}^{2}}}{{{\left( n! \right)}^{2}}}\times \dfrac{\left( kn \right)!}{\left( kn+k \right)!}$
Simplifying the above equation, we get,
$\Rightarrow \dfrac{{{\left( n+1 \right)}^{2}}}{\left( kn+k \right)\left( kn+k-1 \right)......\left( kn+1 \right)}$
$\therefore \left| \dfrac{{{x}_{n}}}{{{x}_{n+1}}} \right|=\dfrac{{{\left( n+1 \right)}^{2}}}{\left( kn+k \right)\left( kn+k-1 \right)......\left( kn+1 \right)}$
Now, we have to find out the value of the limit for different values of k.
Following the same, for $k=1$, the value of the limit is
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{n+1}$
Cancelling the common factors on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\left( n+1 \right)$
Evaluating the limit,
$\therefore \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=+\infty$ for $k=1$
For $k=2$,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{\left( n+1 \right)\left( n+2 \right)}$
Cancelling the common factors on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{{}}}}{\left( n+2 \right)}$
Dividing the numerator and denominator with $n$ on the right-hand side,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( 1+\dfrac{1}{n} \right)}^{{}}}}{\left( 1+\dfrac{2}{n} \right)}$
As $n \to \infty$, $\dfrac{1}{n}\to \infty$. Following the same,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=1$
The value of the limit is equal to one and the result is inconclusive. So, we have to further evaluate the series $\sum\nolimits_{n=1}^{\infty }{\dfrac{{{\left( n! \right)}^{2}}}{\left( 2n \right)!}}$
Expanding the factorial terms, we get,
$\Rightarrow \sum\nolimits_{n=1}^{\infty }{\dfrac{{{\left( n! \right)}^{2}}}{\left( 2n \right)!}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}$
From series and sequences,
$\Rightarrow \dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}=\prod\limits_{p=1}^{n}{\dfrac{p}{n+p}}$
From the above $p\le n$ ,
$\Rightarrow \dfrac{p}{n+p}\le \dfrac{p}{p+p}\le \dfrac{p}{2p}\le \dfrac{1}{2}$
Substituting the value on the right-hand side, we get,
$\Rightarrow \dfrac{n\left( n-1 \right)\left( n-2 \right)....2.1}{2n\left( 2n-1 \right)........\left( n+1 \right)}\le {{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow \sum\nolimits_{n=1}^{\infty }{{{\left( \dfrac{1}{2} \right)}^{n}}=1}$
The above series is convergent, so the given series is also convergent.
Now,
For the value of $k\ge 3$ ,
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\dfrac{{{\left( n+1 \right)}^{2}}}{\left( n+1 \right)\left( n+2 \right)....\left( n+k \right)}$
$\therefore \displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=0$
$\therefore$The given series is convergent for the value $k\ge 2$
Note: One must remember that the multiplication happens to a given number down to the number one and not zero. Factorial of a number in mathematics is the product of all the positive numbers less than or equal to a number.
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