Estimate the series $\sum_{k=0}^\infty(-1)^k(2k+1)^2 z^{(2k+1)^2}$

Question

I have encounter a series of the form $$\sum_{k=0}^\infty(-1)^k(2k+1)^2 z^{(2k+1)^2}, \ z\in [0,1),$$ which basically comes from the derivative of $$\sum_{k=0}^\infty (-1)^k z^{(2k+1)^2}.$$

The later is a bit like the Jacobi theta function which I managed to calculate its limit as $z \to 1$. But is there any effective way to calculate the first one as $z\to 1$? Numerical calculation shows the limit shall be $-1/2$.

Answer 1

New Answer. Following the definition in this Wikipedia article, we define $(\mathrm{C},\alpha)$-summability as follows:

Definition. A sequence $\{a_n\}_{n\geq 0}$ is called $(\mathrm{C},\alpha)$-summable if $\lim_{n\to\infty} S^{\alpha}_n$ converges, where $$ S^{\alpha}_n := \sum_{k=0}^{n} \frac{\binom{n}{k}}{\binom{n+\alpha}{k}} a_k. $$ In such case, the value of $\lim_{n\to\infty} S^{\alpha}_n$ is denoted by $(\mathrm{C},\alpha)\text{-}\sum_{n=0}^{\infty}a_n$.

Then the following version of the abelian theorem holds:

Theorem. Let $(a_n)_{n\geq 0}$ be a s$(\mathrm{C},\alpha)$-summable sequence for some non-negative integer $\alpha$. Then for any real polynomial $\lambda(n)$ such that $\lambda(n)\to\infty$ as $n\to\infty$, we have

$$ \lim_{s \to 0^+} \sum_{n=0}^{\infty} a_n e^{-s\lambda(n)} = (\mathrm{C},\alpha)\text{-}\sum_{n=0}^{\infty} a_n. $$

We first demonstrate how this theorem can be used to prove the existence of OP's limit and find its value. Indeed, it can be shown that $a_n = (-1)^n (2n+1)^2$ is $(\mathrm{C},3)$-summable. Then the above theorem with $\lambda(n) = (2n+1)^2$ shows that

$$ \lim_{s \to 0^+} \sum_{n=0}^{\infty} (-1)^n (2n+1)^2 e^{-s (2n+1)^2} = (\mathrm{C},3)\text{-}\sum_{n=0}^{\infty} (-1)^n (2n+1)^2. $$

On the other hand, this value can also be computed by the above theorem with the choice $\lambda(n) = n$, which leads to

\begin{align*} (\mathrm{C},3)\text{-}\sum_{n=0}^{\infty} (-1)^n (2n+1)^2 &= \lim_{s \to 0^+} \sum_{n=0}^{\infty} (-1)^n (2n+1)^2 e^{-sn} \\ &= \lim_{s \to 0^+} \frac{1 - 6e^{-s} + e^{-2s}}{(1 + e^{-s})^3} \\ &= -\frac{1}{2}. \end{align*}

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Sketch of Proof. The proof of the theorem is a bit lengthy, so we only sketch the idea: if $\deg \lambda = d$, then by substituting $s = t^d$, we get

$$ \sum_{n=0}^{\infty} a_n e^{-t^d \lambda(n)} = \int_{0}^{\infty} \Biggl( \sum_{n=0}^{\infty} S^{\alpha}_n \psi^{\alpha}_{t,n}(u) \Biggr) \left( -\frac{\partial}{\partial u}\right)^{\alpha+1} e^{-t^d \lambda(u/t)} \, \mathrm{d}u, $$

where $\psi_{t,n,\alpha}(u)$ is defined by

$$ \psi^{\alpha}_{t,n}(u) := \binom{n+\alpha}{n} \sum_{j=0}^{\alpha+1} (-1)^j \binom{\alpha+1}{j} \frac{\max\{u - t(n+j),0\}^{\alpha}}{\alpha!}. $$

This function satisfies a very nice property:

$$ \psi^{\alpha}_{t,n}(u) \geq 0 \qquad \text{and} \qquad \frac{u^{\alpha}}{\alpha!}\mathbf{1}_{[0, Nt]}(u) \leq \sum_{n=0}^{N-1} \psi^{\alpha}_{t,n}(u) \leq \frac{u^{\alpha}}{\alpha!} $$

for any $u \geq 0$ and $N \geq 1$. Now let $c > 0$ be the leading coefficient of $\lambda$. Then using the above observation together with a bit of estimates, we can prove that $S^{\alpha}_n \to S$ implies

$$ \lim_{t \to 0^+} \sum_{n=0}^{\infty} a_n e^{-t^d \lambda(n)} = \int_{0}^{\infty} \left( S \frac{u^{\alpha}}{\alpha!} \right) \left( -\frac{d}{d u}\right)^{\alpha+1} e^{-cu^d} \, \mathrm{d}u = S. $$

Answer 2

If we consider (for $x\in\mathbb{R}^+$) $$ f(x) = \sum_{k\geq 0}(-1)^k e^{-(2k+1)^2 x}\stackrel{\text{SBP}}{=}\sum_{k\geq 0}\left[e^{-(4k+1)^2 x}-e^{-(4k+3)^2 x}\right]$$ by invoking the Fourier or Laplace transform we have that $f(x)$ behaves like $\frac{1+x}{2}$ in a right neighbourhood of the origin. This depends on the asymptotic behaviour of the digamma function, i.e. on the asymptotic behaviour of harmonic numbers. It gives $$ \lim_{x\to 0^+} \sum_{k\geq 0}(-1)^k (2k+1)^2 e^{-(2k+1)^2 x} =-\frac{1}{2}$$ and the claim after the substitution $x=-\log z$: $$ \lim_{z\to 1^-}\sum_{k\geq 0}(-1)^k (2k+1)^2 z^{(2k+1)^2} =-\frac{1}{2}.$$