Prove that for every $\vartheta, 0 < \vartheta < 1$, there exists a sequence $\lambda_{n}$ of positive integers and a series $\sum_{n=1}^{\infty} a_{n}$ such that: \begin{align*} (i) & \quad \lambda_{n+1} - \lambda_{n} > (\lambda_{n})^{\vartheta}, \\ (ii) & \quad \lim_{r \to 1-0} \sum_{n=1}^{\infty} a_{n} r^{\lambda_{n}} \text{ exists}, \\ (iii) & \quad \sum_{n=1}^{\infty} a_{n} \text{ is divergent}. \end{align*}
Background:
András Simonovits noted that $\lambda_{n+1} > c\lambda_{n}$ ($c > 1$) and the $(C,1)$ summability together guarantee convergence. Moreover, if we substitute $x = e^{-y}$ in G. H. Hardy, $\textit{Divergent Series}$, Clarendon Press, Oxford, 1949, p. 87, Theorem 114 can be stated as follows: Given $c > 1$ constant and a sequence of natural numbers, such that $\lambda_{n+1} > c\lambda_{n}$ ($c > 1$), $\sum_{n=1}^{\infty} a_{n} r^{\lambda_{n}}$ converges in the unit circle, and its limit exists as $r \to 1^{+}$, then $\sum_{n=1}^{\infty} a_{n}$ is convergent.
So, this theorem, conjectured by Littlewood and proved by Hardy and Littlewood, states that Abel summability and convergence are equivalent notions in the case of sequences satisfying the Hadamard gap condition $\lambda_{n+1} > c\lambda_{n}$ ($c > 1$). Our problem states that the Hadamard gap condition cannot be substituted with a much weaker condition. The participants gave two kinds of generalizations for the problem. Some of them substitute Abel summability with the stronger $(C,1)$ summability; others showed that for any sequence $\lambda_{n}$ satisfying $\liminf (\lambda_{n+1} - \lambda_{n})/\lambda_{n} = 0$, there exists a sequence $\sum_{n=1}^{\infty} a_{n}$ that satisfies the conditions of the problem. This latter statement is also a generalization of the theorem, since if $\lambda_{n} = e^{\sqrt{n}}$, then...
$$\lambda_{n+1} - \lambda_n = e^{\sqrt{n+1}} - e^{\sqrt{n}} = (e^{\sqrt{n+1} - \sqrt{n}} - 1)e^{\sqrt{n}}$$ $$= e^{\sqrt{n}} \left(e^{\frac{1}{\sqrt{n} + \sqrt{n+1}}} - 1\right) = \left(1 + o(1)\right) \frac{e^{\sqrt{n}}}{2\sqrt{n}}$$ And therefore $$\frac{\lambda_{n+1} - \lambda_{n}}{\lambda_{n}} \rightarrow 0, \quad \frac{\lambda_{n+1} - \lambda_{n}}{\lambda_{n}^{\vartheta}} \rightarrow \infty \quad \text{for } \vartheta < 1$$