
$\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} $ is equal to
(A) $\ln x$
(B) $x$
(C) $\dfrac{1}{{\ln x}}$
(D) $\dfrac{1}{x}$
Answer
583.5k+ views
Hint: n the given problem, we have to find the sum of infinite series. The series expansion of the function $f\left( x \right) = {e^x}$ is given by ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \ldots \ldots + \dfrac{{{x^n}}}{{n!}} + \ldots \ldots $. It can be written as ${e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} $. We will use this information in the given problem.
Complete step-by-step answer:
In the given problem, $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} $ is an infinite series. We have to find the sum of this infinite series. We know that the series expansion of the function $f\left( x \right) = {e^x}$ is given by $\Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \ldots \ldots + \dfrac{{{x^n}}}{{n!}} + \ldots \cdots \cdots \left( 1 \right)$.
Equation $\left( 1 \right)$ can be written as ${e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}} \cdots \cdots \left( 2 \right)} $.
Let us replace $x$ by $\ln x$ on both sides of equation $\left( 2 \right)$. So, we can write
${e^{\ln x}} = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} $
$ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}} = {e^{\ln x}} \cdots \cdots \left( 3 \right)} $
Also we know that ${e^{\ln f\left( x \right)}} = f\left( x \right)$. Hence, by using this information we can write ${e^{\ln x}} = x$. So, from equation $\left( 3 \right)$ we can write
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} = x$
Hence, the required sum is $x$ of given infinite series. Hence, option B is correct.
So, the correct answer is “Option B”.
Note: If ${a_1},{a_2},{a_3}, \ldots \ldots ,{a_n}$ is the given sequence then the expression ${a_1} + {a_2} + {a_3} + \ldots \ldots + {a_n} + \ldots $ is called the series. The series is the sum of terms of sequence. The series is called finite or infinite according to the given sequence is finite or infinite. Remember that the series expansion of the function $f\left( x \right) = {e^{ - x}}$ is given by ${e^{ - x}} = 1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - \ldots \ldots $. Series expansion of functions is very useful to find the sum of infinite series in these types of problems.
Complete step-by-step answer:
In the given problem, $\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} $ is an infinite series. We have to find the sum of this infinite series. We know that the series expansion of the function $f\left( x \right) = {e^x}$ is given by $\Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \ldots \ldots + \dfrac{{{x^n}}}{{n!}} + \ldots \cdots \cdots \left( 1 \right)$.
Equation $\left( 1 \right)$ can be written as ${e^x} = \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}} \cdots \cdots \left( 2 \right)} $.
Let us replace $x$ by $\ln x$ on both sides of equation $\left( 2 \right)$. So, we can write
${e^{\ln x}} = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} $
$ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}} = {e^{\ln x}} \cdots \cdots \left( 3 \right)} $
Also we know that ${e^{\ln f\left( x \right)}} = f\left( x \right)$. Hence, by using this information we can write ${e^{\ln x}} = x$. So, from equation $\left( 3 \right)$ we can write
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\ln x} \right)}^n}}}{{n!}}} = x$
Hence, the required sum is $x$ of given infinite series. Hence, option B is correct.
So, the correct answer is “Option B”.
Note: If ${a_1},{a_2},{a_3}, \ldots \ldots ,{a_n}$ is the given sequence then the expression ${a_1} + {a_2} + {a_3} + \ldots \ldots + {a_n} + \ldots $ is called the series. The series is the sum of terms of sequence. The series is called finite or infinite according to the given sequence is finite or infinite. Remember that the series expansion of the function $f\left( x \right) = {e^{ - x}}$ is given by ${e^{ - x}} = 1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - \ldots \ldots $. Series expansion of functions is very useful to find the sum of infinite series in these types of problems.
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