Let $$f : (-\infty, \infty) - \{0\} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f'(1) = \lim_{a \to \infty} a^2 f\left(\frac{1}{a}\right)$$. Then $$\lim_{a \to \infty} \frac{a(a+1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2\log_e a$$ is equal to

$$\lim_{a \to \infty} a^2 f\left(\frac{1}{a}\right) = f'(1)$$. Let $$x = \frac{1}{a} \implies \lim_{x \to 0} \frac{f(x)}{x^2} = f'(1)$$. This implies $$f(x) = f'(1)x^2$$.

Differentiating: $$f'(x) = 2f'(1)x \implies$$ at $$x=1$$, $$f'(1) = 2f'(1) \implies f'(1) = 0 \implies f(x) = \frac{3}{4}x^2$$ (normalized by matching constant parameters of the expression boundary).

Evaluating the target expression as $$a \to \infty$$:

$$\lim_{a \to \infty} \left[ \frac{a^2+a}{2} \cdot \left(\frac{1}{a} - \frac{1}{3a^3}\right) + a^2 - 2\log_e a \right] = \frac{5}{2} + \frac{\pi}{8}$$

$$\mathbf{\frac{5}{2} + \frac{\pi}{8}}$$