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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,

$$\lim_{\alpha\to 0^+} \left[ \lim_{x\to 1^-}\frac{\alpha - \sum_{k=0}^\infty \left( x^{k^2}-x^{(k+\alpha)^2}\right)}{1-x} \right]=\frac16$$

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Mark Fischler
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1 Answers1

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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.

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Mark Viola
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  • @JackD'Aurizio Thank you Jack! – Mark Viola Jan 05 '17 at 02:08
  • Very nice. This reveals that for very small $\alpha$ the behavior as $\x \to 1$ is, to first order driven by the $B_2(\log x) $ term. – Mark Fischler Jan 05 '17 at 19:46
  • Yes. I seem to have a factor of $\alpha$ in the result (i.e., $\alpha/6$) before letting $\alpha \to 0$. So, I don't believe that the limit as $\alpha\to 0$, as written, is $1/6$, but rather is $0$. – Mark Viola Jan 05 '17 at 20:23