If $$0 < a, b < 1$$, and $$\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4}$$, then the value of $$(a+b) - \left(\frac{a^2 + b^2}{2}\right) + \left(\frac{a^3 + b^3}{3}\right) - \left(\frac{a^4 + b^4}{4}\right) + \ldots$$ is:

We are given $$0 < a, b < 1$$ and $$\tan^{-1}a + \tan^{-1}b = \dfrac{\pi}{4}$$.

Using the addition formula, $$\tan\left(\tan^{-1}a + \tan^{-1}b\right) = \dfrac{a + b}{1 - ab} = \tan\dfrac{\pi}{4} = 1$$. Therefore $$a + b = 1 - ab$$, which gives $$a + b + ab = 1$$.

The given series is $$(a + b) - \dfrac{a^2 + b^2}{2} + \dfrac{a^3 + b^3}{3} - \dfrac{a^4 + b^4}{4} + \cdots$$

This can be written as $$\displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{a^k + b^k}{k} = \displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{a^k}{k} + \displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{b^k}{k}$$.

Each sum is the Maclaurin series for $$\ln(1 + x)$$ evaluated at $$x = a$$ and $$x = b$$ respectively (valid since $$0 < a, b < 1$$). So the series equals $$\ln(1 + a) + \ln(1 + b) = \ln\big((1 + a)(1 + b)\big)$$.

Expanding: $$(1 + a)(1 + b) = 1 + a + b + ab = 1 + 1 = 2$$ (using $$a + b + ab = 1$$).

Therefore the value of the series is $$\ln 2 = \log_e 2$$.