Question

Given, $(2/1).(1/3)+(3/2).(1/9)+(4/3).(1/27)+(5/4).(2/81)+........$ is equal to
A.${}^{1}/{}_{2}+{{\log }_{e}}(3/2)$
B.${{\log }_{e}}3-{{\log }_{e}}2$
C.$\log 6$
D.${{\log }_{e}}2-{{\log }_{e}}3$

Answer 1

To solve this problem, first observe the given series and then try to find out the ${{n}^{th}}$ terms of the series and after that apply the summation on that term and try to simplify it using arithmetic operations, after simplifying the summation you will get your required answer.

Complete answer:
Before solving the given question, let’s understand the concept of series:
A series can only be defined as the sum of various numbers, or sequences. Series may or may not end depending on the sequence of events.
The series types are as follows:
A series of Geometry
Harmonic series
Power Series
Interchange Series
Exponent series
A geometric chain can be defined as a series with a continuous balance between successive terms.
A harmonic series can be defined as a series containing a series of terms that replicate the concepts of a mathematical series.
A power chain can be defined as a series that can be thought of as a polynomial with an unlimited number of words.
Now, we will try to understand the Euler number:
$e$ is a real number that is called an Euler's number (another name also known as Napier's constant). This is named after a great mathematician named Leonhard Euler. As you have seen before, it has been mentioned under log activity and is known as the basis for logarithmic activity. This is used not only in Mathematics but also in Physics.
Decimal representation of the Euler’s number is a non terminating and a non repeating number. Non terminating means that the numbers in the decimal will go on forever and non repeating means that it does not settle down to any pattern.
Now, according to the given question:
Let, $S=(2/1).(1/3)+(3/2).(1/9)+(4/3).(1/27)+(5/4).(2/81)+........$
After observing the series, the ${{n}^{th}}$ term is given as:
${{T}_{n}}=\left( \dfrac{n+1}{n} \right).\left( \dfrac{1}{{{3}^{n}}} \right)$
Now simplifying it as follows:
$\Rightarrow {{T}_{n}}=\left( \dfrac{n}{n}+\dfrac{1}{n} \right).\left( \dfrac{1}{{{3}^{n}}} \right)$
$\Rightarrow {{T}_{n}}=\left( 1+\dfrac{1}{n} \right).\left( \dfrac{1}{{{3}^{n}}} \right)$
Adding summation in this term:
$\Rightarrow \sum\limits_{n=1}^{\infty }{{{T}_{n}}}$
$\Rightarrow \sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{{{3}^{n}}} \right)}+\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{n{{3}^{n}}} \right)}$
Dividing both the terms with $(1-{}^{1}/{}_{3})$ , we will get as:
\Rightarrow \left[ \left( {}^{1}/{}_{3} \right)/\left( 1-{}^{1}/{}_{3} \right) \right]+\left\\{ -{{\log }_{e}}\left( 1-{}^{1}/{}_{3} \right) \right\\}
$\Rightarrow \dfrac{1}{2}+{{\log }_{e}}\left( \dfrac{3}{2} \right)$

Hence, the correct option from all above given options is $A$.

Note:
Logarithms make our life easier and logarithmic functions are used in many fields. One of the examples of logarithmic scale in real life is the Richter scale that is used for earthquakes. The largest earthquake ever recorded was of magnitude $9.5$ in Chile on a $1,000$ mile long fault line.