Limits involving Euler-Mascheroni Constant!

This looks quite messy, so let's fiddle about and see what happens.

I start with simplifying this: $S=\sum_{i=1}^{2n} \dfrac{(-1)^{i-1}}{i} {2n \choose i}$

So we consider:

$(1-x)^{2n}=1-\sum_{i=1}^{2n} {(-1)^{i-1}} {2n \choose i}x^{i}$

If we divide by $x$ and integrate we can get it to match what we want:

$S=\sum_{i=1}^{2n} \dfrac{(-1)^{i-1}}{i} {2n \choose i}=\int_{0}^{1}\dfrac{1-(1-x)^{2n}}{x}dx$

By making a $u=1-x$ sub we can simplify further:

$S=\int_{0}^{1}\frac{1-u^{2n}}{1-u}du=\int_{0}^{1}\sum\limits_{k=0}^{2n-1}u^k\,dx=H_{2n}$

$L = \lim_{n \to \infty} \left( H_{2n} - \ln(8n) \right)= \lim_{n \to \infty} \left( H_{2n} - \ln(2n) \right)-\ln(4)$

$\therefore L=γ-\ln(4) \implies A=\boxed{4}$