Observe the following lists
List - I List - II
A) $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \dots \infty$ 1) $\frac{3}{2} \log 2$
B) $\frac{1}{5} + \frac{1}{2 . 5^{2}} + \frac{1}{3 . 5^{3}} + \dots \dots \infty$ 2) $\log_{e} 2$
C) $\frac{1}{n + 1} + \frac{1}{2 {(n + 1)}^{2}} + \frac{1}{3 {(n + 1)}^{3}} . \dots \infty$ 3) $\log_{e} (\frac{4}{5})$
D) $1 + \frac{1}{3 . 3^{2}} + \frac{1}{5 . 3^{4}} \dots \dots \infty$ 4) $\log_{e} (\frac{5}{4})$
5) $\log_{e} (1 + \frac{1}{n})$
The correct match for List - I from List - II is

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A B C D

2 4 1 5

A B C D

4 2 5 3

A B C D

3 2 5 4

A B C D

2 4 5 1

answer is D.

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Detailed Solution

(A) $\begin{array}{rcl} 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + . . . . . . . \infty \\ = & \log (1 + 1) = \log_{e}^{2} \\ (A) \to (2) \end{array}$

(B) $\begin{array}{rcl} \frac{1}{5} + \frac{{(\frac{1}{5})}^{2}}{2} + \frac{{(\frac{1}{5})}^{3}}{3} + . . . . . . \infty \\ = & - \log (1 - \frac{1}{5}) \\ = & \log (\frac{5}{4}) \\ (B) \to (4) \end{array}$

(C) $\begin{array}{rcl} \frac{1}{n + 1} + {\frac{1}{2 (n + 1)}}^{2} + . . . . . . \\ = & - \log (1 - \frac{1}{n + 1}) \\ = & \log (\frac{n + 1}{n}) = \log (1 + \frac{1}{n}) \\ (C) & \to & (5) \end{array}$

(D) $\begin{array}{rcl} 1 + \frac{{(\frac{1}{3})}^{2}}{3} + \frac{{(\frac{1}{3})}^{4}}{5} + . . . . . \\ = & 3 [\frac{1}{3} + \frac{{(\frac{1}{3})}^{3}}{2} + \frac{{(\frac{1}{5})}^{5}}{5} + . . . . .] \\ = & 3 [\frac{1}{2} \log (\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}})] \\ = & \frac{3}{2} \log (2) \\ (D) & \to & (1) \end{array}$