\[{\log _e}2 + {\log _e}(1 + \dfrac{1}{2}) + {\log _e}(1 + \dfrac{1}{3}) + .......{\log _e}(1 + \dfrac{1}{{n - 1}})\] is equal to
A) \[{\log _e}1\]
B) \[{\log _e}n\]
C) \[{\log _e}(1 + n)\]
D) \[{\log _e}(1 - n)\]

Hint: in this question, we have to find the sum of the given infinite series. In order to solve this first Rearrange the given series to find standard pattern of the series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.



Formula Used:\[{\log _e}(\dfrac{a}{b}) = {\log _e}(a) - {\log _e}(b)\]
Some expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...\]
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]



Complete step by step solution:We have \[{\log _e}2 + {\log _e}(1 + \dfrac{1}{2}) + {\log _e}(1 + \dfrac{1}{3}) + .......{\log _e}(1 + \dfrac{1}{{n - 1}})\]
Now we have to rearrange the logarithm series in order to find standard form of logarithm series.
\[{\log _e}2 + {\log _e}(1 + \dfrac{1}{2}) + {\log _e}(1 + \dfrac{1}{3}) + .......{\log _e}(1 + \dfrac{1}{{n - 1}}) = {\log _e}2 + {\log _e}(\dfrac{3}{2}) + {\log _e}(\dfrac{4}{3}) + .......{\log _e}(\dfrac{n}{{n - 1}})\] (i)
We know that \[{\log _e}(\dfrac{a}{b}) = {\log _e}(a) - {\log _e}(b)\]
After apply this in equation (i) we get
\[ = {\log _e}(2) + {\log _e}(3) - {\log _e}(2) + {\log _e}(4) - {\log _e}(3) + ....{\log _e}(n) - {\log _e}(n - 1) = {\log _e}(n)\]



Option ‘A’ is correct



Note: Here we have to rearrange the series in order to get standard pattern of series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.
In this question after rearrangement we found that series is written as a standard logarithm series. After getting standard series we have to apply the \[{\log _e}(\dfrac{a}{b}) = {\log _e}(a) - {\log _e}(b)\] formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.