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
Verified605.4k+ views
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.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE
Draw a labelled diagram of the human male reproductive class 12 biology CBSE
Describe the experiment to compare the emf of two cells class 12 physics CBSE
What is standard hydrogen electrode
What is conventional current and electric current class 12 physics CBSE
2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE
Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE
Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?