Question

How do you use the ratio test to test the convergence of the series $\sum {\dfrac{{n!}}{{{n^n}}}}$ from $n = 1$ to infinity?

Answer 1

First let us know what the ratio test states. So, the ratio test says that a series $\sum {{x_n}}$is convergent if $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| < 1$ and is divergent if $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| > 1$ and the test fails if $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = 1$. So, to solve this problem, we are to first find the ${x_{n + 1}}$ term, that is the term of the series that is $n + 1$th term. Then we will find, $\left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right|$ and then $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right|$. Then applying the ratio test accordingly we can find, whether, $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right|$ is less than $1$ or greater than $1$, that is, whether the series is convergent or divergent respectively.

Complete step by step solution:
Given, is $\sum\limits_{n = 1}^\infty {\dfrac{{n!}}{{{n^n}}}}$.
The ${n^{th}}$ term of the series is, ${x_n} = \dfrac{{n!}}{{{n^n}}}$.
Therefore, the ${\left( {n + 1} \right)^{th}}$ term of the series is, ${x_{n + 1}} = \dfrac{{(n + 1)!}}{{{{(n + 1)}^{n + 1}}}}$.
Now, evaluating the ratio, $\left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \dfrac{{\dfrac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}}}{{\dfrac{{n!}}{{{n^n}}}}}$.
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \dfrac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}.\dfrac{{{n^n}}}{{n!}}$
We know, $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1$
Using, this property, we get,
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \dfrac{{\left( {n + 1} \right).n!}}{{{{\left( {n + 1} \right)}^n}.\left( {n + 1} \right)}}.\dfrac{{{n^n}}}{{n!}}$
Now, cancelling the common terms between the numerator and denominator, we get,
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \dfrac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}}}$
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = {\left( {\dfrac{n}{{n + 1}}} \right)^n}$
Now, dividing both numerator and denominator by ${n^n}$, we get,
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = {\left( {\dfrac{{\dfrac{n}{n}}}{{\dfrac{n}{n} + \dfrac{1}{n}}}} \right)^n}$
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = {\left( {\dfrac{1}{{1 + \dfrac{1}{n}}}} \right)^n}$
$\Rightarrow \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \dfrac{1}{{{{\left( {1 + \dfrac{1}{n}} \right)}^n}}}$
Now, taking limit of the ratio, we get,
$\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{{{{\left( {1 + \dfrac{1}{n}} \right)}^n}}}} \right)$
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \left( {\dfrac{1}{{\mathop {\lim }\limits_{n \to \infty } {{\left( {1 + \dfrac{1}{n}} \right)}^n}}}} \right)$
Now, we know, $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \dfrac{1}{n}} \right)^n} = e$
Therefore, using this property, we can write,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \left( {\dfrac{1}{e}} \right)$
Now, we know, $2 < e < 3$.
That is clearly, $e > 1$
$\Rightarrow \dfrac{1}{e} < 1$
Therefore, we can say,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = \left( {\dfrac{1}{e}} \right) < 1$
By applying the ratio test, we can clearly say, the series is convergent.

Note: The ratio test is one of the ways to test the convergence of a series. It is preferable to use the test if the series has terms in the form of fractions. Other than the ratio test, there is also the Cauchy’s root test, in which a series $\sum {{x_n}}$is convergent if $\mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {{x_n}} \right)}^{\dfrac{1}{n}}}} \right| < 1$ and is divergent if $\mathop {\lim }\limits_{n \to \infty } \left| {\left| {{{\left( {{x_n}} \right)}^{\dfrac{1}{n}}}} \right|} \right| > 1$ and the test fails if $\mathop {\lim }\limits_{n \to \infty } \left| {\left| {{{\left( {{x_n}} \right)}^{\dfrac{1}{n}}}} \right|} \right| = 1$.