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Question: The value of \[\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \r...

The value of log⁡3+(log⁡3)33!+(log⁡3)55!+.....+∞\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \right)}^5}}}{{5!}} + ..... + \infty .

Explanation

Solution

In the given questions , we have given with a series of sinh⁡(x)\sinh \left( x \right) ( i.e. sine hyperbolic ) which is sinh⁡(x)=ex−e−x2\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2} , we will be using the expansion of the Taylor series of sinh⁡x=x+x33!+x55!+.....+∞\sinh x = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ..... + \infty , on equating the expansion with the expression of sinh⁡(x)\sinh \left( x \right) we will get the required answer .

Complete step by step answer:
The Taylor series of a function is an infinite sum of all the terms that are expressed in terms of the function's derivatives at a single point . For most common functions, the function and the sum of its Taylor series are equal near this point.
Given: log⁡3+(log⁡3)33!+(log⁡3)55!+.....+∞\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \right)}^5}}}{{5!}} + ..... + \infty .
Now, using the Taylor expansion for sinh⁡(x)\sinh \left( x \right) we get,
sinh⁡x=x+x33!+x55!+.....+∞\sinh x = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ..... + \infty
On comparing given expression with the Taylor expansion, we have x=log⁡3x = \log 3 ,
Now using the general exponential term for Taylor expansion, we have,
sinh⁡(x)=ex−e−x2\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2} .
Now, putting x=log⁡3x = \log 3 in the above expression, we get
=elog⁡3−e−log⁡32= \dfrac{{{e^{\log 3}} - {e^{ - \log 3}}}}{2}
Using the property of exponent i.e. elog⁡x=x{e^{\log x}} = x , we get
=3−e−log⁡32= \dfrac{{3 - {e^{ - \log 3}}}}{2}
Now, using the property of −log⁡x=log⁡1x - \log x = \log \dfrac{1}{x} , we get
=3−132= \dfrac{{3 - \dfrac{1}{3}}}{2}
On solving we get ,
=832= \dfrac{{\dfrac{8}{3}}}{2}
On solving further we get ,
=43= \dfrac{4}{3}

Note: The given expression can not be solved directly , as it is a series so you have to find out a common term for the given series . Moreover, some series are predefined as in the question we have series of sinh⁡(x)\sinh \left( x \right) ( i.e. sine hyperbolic ) . Also , to solve questions related to log⁡\log you must have prior knowledge about the properties of log⁡\log and the same goes with the ee ( exponent ) . Also, check whether the series is up to infinity (∞)\left( \infty \right) or not , as then the question will be related to AP or GP .