How fast does $\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)$ go to $\alpha$? In this question: $\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)$ it is established that 
$$\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right) = \alpha$$
 for all $\alpha > 0$. 
I'd like to know how fast it approaches $\alpha$. In particular, to show that the rate of approach is such that for all $\alpha$ 
$$\lim_{x\to 1^-}\frac{\alpha -  \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)}{1-x}$$
exists, and that when $\alpha$ is very small,
$$\lim_{\alpha\to 0^+} \left[ \lim_{x\to 1^-}\frac{\alpha -  \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)}{1-x}
\right]=\frac16$$
 A: We use the same approach as in THIS ANSWER.  From the Euler Maclaurin Summation Formula, we have
$$\begin{align}
\sum_{k=1}^K (x^{k^2}-x^{(k+\alpha)^2})&=\int_0^K (e^{y^2\log(x)}-e^{(y+\alpha)^2\log(x)})\,dy\\\\
&+\frac{e^{-K^2|\log(x)|}-e^{-(K+\alpha)^2|\log(x)|}-(1-e^{-\alpha^2|\log(x)|})}2  \\\\
&+\frac{B_2\,\log(x)}{2!}\left(2Ke^{-K^2|\log(x)|}-2(K+\alpha)e^{-(K+\alpha)^2|\log(x)|}-(0-2\alpha e^{-\alpha^2|\log(x)|})\right)\\\\
&+\log^2(x)\int_0^K \left(4y^2e^{-y^2|\log(x)|}-4(y+\alpha)^2e^{-(y+\alpha)^2|\log(x)|}\right)P_2(y)\,dy \\\\
&=\int_0^\alpha e^{-y^2|\log(x)|}\,dy-\int_K^{K+\alpha}e^{-y^2|\log(x)|}\,dy\\\\
&+\frac{e^{-K^2|\log(x)|}-e^{-(K+\alpha)^2|\log(x)|}-(1-e^{-\alpha^2|\log(x)|})}2  \\\\
&+\frac{\log(x)}{12}\left(2Ke^{-K^2|\log(x)|}-2(K+\alpha)e^{-(K+\alpha)^2|\log(x)|}-(0-2\alpha e^{-\alpha^2|\log(x)|})\right)\\\\
&+\log^2(x)\int_0^K \left(4y^2e^{-y^2|\log(x)|}-4(y+\alpha)^2e^{-(y+\alpha)^2|\log(x)|}\right)P_2(y)\,dy \tag 1\\\\
\end{align}$$
Letting $K\to \infty$ in $(1)$ yields
$$\begin{align}
\sum_{k=1}^K (x^{k^2}-x^{(k+\alpha)^2})&=\int_0^\alpha e^{-y^2|\log(x)|}\,dy-\frac{1-e^{-\alpha^2|\log(x)|}}2  \\\\
&+\frac{\alpha \log(x)}{6}e^{-\alpha^2|\log(x)|}\\\\
&+\log^2(x)\int_0^\infty \left(4y^2e^{-y^2|\log(x)|}-4(y+\alpha)^2e^{-(y+\alpha)^2|\log(x)|}\right)P_2(y)\,dy \tag 2\\\\
\end{align}$$

Next, we have the following limits.
LIMIT $1$:
$$\lim_{x\to 1^-}\frac{\alpha -\int_0^\alpha e^{-y^2|\log(x)|}}{1-x}=-\lim_{x\to 1^-}\int_0^\alpha \frac{y^2}{x}e^{-y^2|\log(x)|}\,dy=-\frac12\alpha^2 \tag 3$$

LIMIT $2$:
$$\lim_{x\to 1^-} \frac{1-e^{-\alpha^2|\log(x)|}}{2(1-x)}=-\lim_{x\to 1^-} \frac{\alpha^2e^{-\alpha^2|\log(x)|}}{2x}=-\frac12 \alpha^2 \tag 4$$

LIMIT $3$:
$$\lim_{x\to 1^-}\frac{\frac{\log(x)}{6}e^{-\alpha^2|\log(x)|}}{1-x}=-\frac16 \tag 5$$

LIMIT $4$:
$$\lim_{x\to 1^-} \frac{\log^2(x)}{1-x}\int_0^\infty \left(4y^2e^{-y^2|\log(x)|}-4(y+\alpha)^2e^{-(y+\alpha)^2|\log(x)|}\right)P_2(y)\,dy =0 \tag 6$$

Putting $(3)-(6)$ together yields the coveted limit.
