Convergence of $\sum_{k=0}^{\infty} a^k\cdot b^{k^2}$

Question

Can we say anything about convergence of a series of the form

$$ \begin{aligned} \Phi = \sum_{k=0}^{\infty} a^k\!\cdot\! b^{k^2} \end{aligned} $$

assuming $0\leq a,b\leq 1$ are constants? Is there a way to compute closed-form expression for $\Phi$?

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I tried applying Geometric Series summation formula for partial sums $\Phi_N = \sum_{k=0}^{N} a^k\!\cdot\! b^{k^2}\!\!, \ N\in \mathbb{N}^+\!,$ hoping to take limit as $N\to \infty$ afterwards, but could not get any closed-form expressions.

! If it is any help, $a$ and $b$ are intrinsically connected and expressible as $$ a = \varphi\left(\sqrt{\,l\cdot c^\phantom{x}\!}\right), \quad b = \varphi\left(\,{l}\,\right), \quad \varphi\left(x\right) := \exp \left(\varepsilon^2x^2\right), $$ where $ l,c,\varepsilon >0$ are constants.

Answer 1

The series converges if either of $a, b$ are $<1$. If $a=b=1$ the series diverges.

Proof:

• If $a<1$, then since $b\leq 1$ we have $a^k b^{k^2} \leq a^k$, so $\sum_k a^k b^{k^2} \leq \sum_k a^k$, which is a convergent geometric series.
• If $b<1$, then because $a\leq 1$ we have $a^k b^{k^2}\leq b^{k^2}$, and because $k^2\geq k$ for integers $k$ we have $b^{k^2} \leq b^k$. So $\sum_k a^k b^{k^2} \leq \sum_k b^k$ which is again a convergent geometric series.
• If $a=b=1$, then $a^k b^{k^2} = 1$, so the series is $\sum_k 1$ which diverges.

Evaluating this sum is much tougher. Wolfram Alpha is able to do the case $a=b$, giving the result in terms of special function "elliptic theta".

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Also, note that if you define $a, b$ as in the box, with positive $l,c,\epsilon$, they are in fact $>1$ (because $\exp(x)>1$ for $x>0$). In the more general case where you allow $a,b>1$, you get:

• Convergence if $b<1$, or if $b=1$ and $a\leq 1$
• Divergence if $b>1$, or if $b=1$ and $a\geq 1$

The $b=1$ cases are just geometric series. For $b\neq 1$, an easy way to see the convergence is to rewrite the sum as $\sum_k b^{k^2 + \log_b(a) k}$. For large $k$, the exponent $k^2 +\log_b(a)k$ is bigger than $k$. Thus, the stated convergence properties hold by comparison to the geometric series $\sum_k b^k$, which converges iff $b<1$.