Step-by-step Solution:

We need to find the coefficient of \( x^{100} \) in the expansion of \[ \log _e\left(\frac{1+x}{1+x^2}\right) \] for \( |x| < 1 \).
Step 1: Expansion of \( \log(1+x) \) and \( \log(1+x^2) \)
Using the Taylor series expansion: \[ \log(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \frac{y^4}{4} + \dots \] we expand both parts separately.
Expansion of \( \log(1 + x) \): \[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \] Expansion of \( \log(1 + x^2) \): \[ \log(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \dots \] Step 2: Compute the Difference Now, \[ \log \left( \frac{1+x}{1+x^2} \right) = \log(1+x) - \log(1+x^2) \] Substituting the expansions: \[ (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots) - (x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \dots) \] Rearrange: \[ x - x^2 - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^4}{2} - \frac{x^6}{3} + \frac{x^6}{3} - \dots \] \[ = x - \frac{3}{2} x^2 + \frac{x^3}{3} + \frac{x^4}{4} - \frac{x^6}{3} + \dots \] Step 3: Coefficient of \( x^{100} \)
Observing the pattern, the general term in the expansion follows: \[ \frac{(-1)^{k+1}}{k} x^k \quad \text{for odd } k \] \[ \frac{(-1)^k}{k} x^k \quad \text{for even } k \] For \( x^{100} \), the coefficient is: \[ \frac{(-1)^{100}}{100} = \frac{1}{100} = 0.01 \]