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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.

Mark Fischler
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  • From $\sum_{n=1}^\infty (n+a)^{-s} \Gamma(s/2) = 2\int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{- (n+a)^2 x^2}dx$ and inverse Mellin/Laplace we get that $\sum_{n=1}^\infty e^{- (n+a)^2 x^2} \sim \frac{\sqrt{\pi}}{x}$ as $x \to 0$ . Otherwise, the Poisson summation applied to $\sum_{n=1}^\infty e^{- (n+a)^2 x}$ should give the same result. – reuns Jan 03 '17 at 19:21
  • I don't understand how that helps with the first statement. The exponentiation involved in the question is for a number just below 1 raised to what quickly become high powers, so how is the pole at $x=0$ important? – Mark Fischler Jan 03 '17 at 19:26
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    https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_tauberian_theorem – Jack D'Aurizio Jan 03 '17 at 19:28

2 Answers2

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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.

Jack D'Aurizio
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    Nice answer. (+1) – Olivier Oloa Jan 03 '17 at 19:48
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    I really like the idea of introducing the approximation-to-the-identity. (+1) – Sangchul Lee Jan 03 '17 at 20:01
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    Happy New Year Jack. Great solution. (+1) I took a different approach and applied the Euler-Maclaurin Summation Formula. -Mark – Mark Viola Jan 03 '17 at 20:27
  • I don't see that this gives $\frac16$ for the second limit (but I had written it wrong, so unless you were a mind reader, you probably could not have attacked that part of the problem. At any rate, I would imagine your technique could be brought to bear, but I am not strong enough to see how. I will ask the second part in a separate question. – Mark Fischler Jan 05 '17 at 00:03
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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!

Mark Viola
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  • The asymptotical expansion $$\psi(x)=\log(x)-\frac{1}{2x}-\sum_{n=1}^{+\infty}\frac{B_{2n}}{2n x^{2n}}$$ is usually derived from the EMC formula, so I guess our solutions are not that different. (+1) and happy new year. – Jack D'Aurizio Jan 03 '17 at 22:31
  • I assume here that $P_1(x) = (x-\lfloor x \rfloor) - 1$. I like this one better than Jack's just because I could not work out his computation of the last series via the digamma function, while I could follow your reasoning. – Mark Fischler Jan 04 '17 at 23:45
  • @MarkFischler Mark, you are correct regarding $P_1(x)$. And Happy New Year! - Mark – Mark Viola Jan 05 '17 at 00:29