The nth term of `log_(e)2` is

Options

A. `((-1)^(n-1))/(n.2^(n))`
B. `((-1)^(n-1))/(n)`
C. `((-1)^(n-1))/(n.3^(n))`
D. `((-1)^(n-1))/(n.4^(n))`

Answer

To find the nth term of \( \log_e 2 \), we can use the Taylor series expansion of \( \log_e(1 + x) \). The series expansion is given by:

\[ \log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

In our case, we can express \( \log_e 2 \) as:

\[ \log_e 2 = \log_e(1 + 1) = \log_e(1 + x) \quad \text{where } x = 1 \]

Now substituting \( x = 1 \) into the series:

\[ \log_e(1 + 1) = 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4} + \ldots \]

This simplifies to:

\[ \log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]

To find the nth term of this series, we observe the pattern. The nth term can be expressed as:

\[ T_n = (-1)^{n-1} \frac{1}{n} \]

This is because the series alternates in sign, and the denominators are simply the natural numbers.

Thus, we can conclude that the nth term of \( \log_e 2 \) is:

\[ T_n = (-1)^{n-1} \frac{1}{n} \]

Now, we need to check the options provided in the question. The correct option that matches our derived nth term is:

\[ \text{Option 2: } (-1)^{n-1} \frac{1}{n} \]