$\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)$ Has anybody seen (or can anybody come up with) a proof that 
 $$\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right) = \alpha$$ 
for all $\alpha > 0$?
And also that 
 $$\lim_{\alpha\to 0^+} \left[ \lim_{x\to 1^-}\frac{\alpha -  \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)}{1-x}
\right]=\frac16$$
I thought I had a proof of the first limit but it was flawed.
EDIT I had written the wrong expression for the second of the limits.  It is now corrected.
 A: If we set $x=e^{-t}$ we have to prove that
$$ \lim_{t\to 0^+}\sum_{k\geq 0}\left(e^{-tk^2}-e^{-t(k+\alpha)^2}\right) = \alpha.$$
As a function of $t$, the above series is regular enough to make us able to state:
$$ \lim_{t\to 0^+}\sum_{k\geq 0}\left(e^{-tk^2}-e^{-t(k+\alpha)^2}\right) = \lim_{m\to +\infty}\int_{0}^{+\infty}me^{-mt}\sum_{k\geq 0}\left(e^{-tk^2}-e^{-t(k+\alpha)^2}\right)\,dt$$
(we use $me^{-mt}$ as an approximation of the Dirac delta distribution) and the original limit is converted into:
$$ \lim_{m\to +\infty}\sum_{k\geq 0}\left(\frac{m}{m+k^2}-\frac{m}{m+(k+\alpha)^2}\right)=\lim_{m\to +\infty}\sum_{k\geq 0}\frac{m(2\alpha k+\alpha^2)}{(m+k^2)(m+(k+\alpha)^2)}$$
Now the last series can be computed through the digamma function and the limit as $m\to +\infty$ is exactly $\alpha$ as wanted, since:
$$\frac{i}{2}\left[\psi(\alpha-im)-\psi(\alpha+im)\right] = \frac{\pi}{2}+\frac{\tfrac{1}{2}-\alpha}{m}+O\left(\frac{1}{m^3}\right). $$
The second question is essentially equivalent to finding an extra term in the previous asymptotic expansion.
A: From the Euler-Maclaurin Summation Formula, we have
$$\begin{align}
\sum_{k=1}^K (x^{k^2}-x^{(k+\alpha)^2})&=\int_0^K (e^{y^2\log(x)}-e^{(y+\alpha)^2\log(x)})\,dy\\\\
&+\frac{e^{-K^2|\log(x)|}-e^{-(K+\alpha)^2|\log(x)|}-(1-e^{-\alpha^2|\log(x)|})}2  \\\\
&+\log(x)\int_0^K \left(2ye^{-y^2|\log(x)|}-2(y+\alpha)e^{-(y+\alpha)^2|\log(x)|}\right)P_1(y)\,dy \\\\
&=\int_0^\alpha e^{-y^2|\log(x)|}\,dy-\int_K^{K+\alpha}e^{-y^2|\log(x)|}\,dy\\\\
&+\frac{e^{-K^2|\log(x)|}-e^{-(K+\alpha)^2|\log(x)|}-(1-e^{-\alpha^2|\log(x)|})}2  \\\\
&+\log(x)\int_0^K \left(2ye^{-y^2|\log(x)|}-2(y+\alpha)e^{-(y+\alpha)^2|\log(x)|}\right)P_1(y)\,dy \tag 1\\\\
\end{align}$$
Taking the limit as $K\to \infty$ in $(1)$ reveals
$$\begin{align}
\sum_{k=1}^\infty (x^{k^2}-x^{(k+\alpha)^2})&=\int_0^\alpha e^{-y^2|\log(x)|}\,dy
-\frac{1-e^{-\alpha^2|\log(x)|}}2  \\\\
&+\log(x)\int_0^\infty \left(2ye^{-y^2|\log(x)|}-2(y+\alpha)e^{-(y+\alpha)^2|\log(x)|}\right)P_1(y)\,dy  \tag 2
\end{align}$$
Taking the limit as $x\to 1^-$ in $(2)$ yields the coveted result
$$\lim_{x\to ^-1}\sum_{k=0}^\infty (x^{k^2}-x^{(k+\alpha)^2})=\alpha$$
as was to be shown!
