The value of $\sum\limits_{n = 2}^\infty {\ln \left( {1 - \dfrac{1}{{{n^2}}}} \right)} $ equals
A.$ - \ln 3$
B.$0$
C.$ - \ln 2$
D.$ - \ln 5$

Hint: An exponent that is written in a special way is known as a logarithm. Logarithm functions are just opposite or inverse of exponential functions. We can easily express any exponential function in a logarithm form. Similarly, all the logarithm functions can be easily rewritten in exponential form. $\sum\limits_{}^{} {} $ is a mathematical letter whose meaning is ‘the sum of’. In order to solve this equation, we will have to use the properties of sigma expansion as well as of logarithm.

Complete step by step solution:
Given is $\sum\limits_{n = 2}^\infty {\ln \left( {1 - \dfrac{1}{{{n^2}}}} \right)} $
We know that according to sigma expansion and logarithm properties,
$ \Rightarrow \ln \left( {1 - \dfrac{1}{{{n^2}}}} \right) = \ln \left( {n + 1} \right) + \ln \left( {n - 1} \right) - 2\ln n$
We are given that $n = 2$ so,
$
   = \ln \left( {2 + 1} \right) + \ln \left( {2 - 1} \right) - 2\ln 2 \\
   = \ln 3 + \ln 1 - 2\ln 2 + \ln 4 + \ln 2 \\
   = {{\ln 3}} + {{\ln 1}} - \ln 2 - {{\ln 2}} + {{\ln 4}} + {{\ln 2}} - {{\ln 3}} - {{\ln 3}} \\
   = - \ln 2 \\
 $
Therefore, the value of $\sum\limits_{n = 2}^\infty {\ln \left( {1 - \dfrac{1}{{{n^2}}}} \right)} $ is $ - \ln 2$.

Hence, the correct option is (C).

Note:
This problem and similar to these can very easily be solved by making use of different logarithm properties. Students should keep in mind the properties of logarithmic functions. Logarithms are useful when we want to work with large numbers. Logarithm has many uses in real life, such as in electronics, acoustics, earthquake analysis and population prediction. When the base of common logarithm is $10$ then, the base of a natural logarithm is number $e$.