$\lim_{x\to 1^-} \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)$

Question

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.

Answer 1

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!

Answer 2

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.