Lacunary functions and sums of reciprocals Let $\Lambda=\left\{ \lambda_{n}\right\} _{n=0}^{\infty}$ be an infinite, strictly increasing sequence of non-negative integers. I say that $\Lambda$ is reciprocal-summable if: $$\sum_{\lambda\in\Lambda\backslash\left\{ 0\right\} }\frac{1}{\lambda}<\infty$$
Now, in a slight abuse of terminology, I'm going to use the term “lacunary” to refer only to those holomorphic functions on $\mathbb{D}$ for which the unit circle is a natural boundary (such as $\sum_{n=0}^{\infty}z^{2^{n}}$). The only functions I'm concerned with are those for which the power series coefficients are $0$s and $1$s. 
I've been reading up on the known criteria (and converses) for when a function is lacunary (results of Fabry, Hadamard, Pólya, etc.). However, many (if not most) of them try to go from the most general point of view they can find, and so, I can't seem to find the exact details / answers that I'm looking for.
The claims which I would like to have answered (either partially or entirely) and/or be pointed toward a counterexample of are as follows:
I. If $\Lambda$
  is reciprocal summable, then $\sum_{n=0}^{\infty}z^{\lambda_{n}}$
  is lacunary.
II. If $\Lambda$
  is not reciprocal-summable, then $\sum_{n=0}^{\infty}z^{\lambda_{n}}$
  is not lacunary.
If any of the results (or predecessors thereof) of Hadamard, Fabry, Pólya (etc.) imply one or more of these claims, an explanation of how they do so would be much appreciated.
Thanks in advance!
 A: A counterexample for Claim II is: $$f(z)=\frac{1}{1-z}-\sum_{n=0}^\infty z^{2^n}$$
On the one hand, $f(z)$ is a difference between a rational and a lacunary function and is therefore lacunary. On the other hand, $$\frac{1}{1-z}=1+z+z^2+z^3+\ldots,$$
so most of the coefficients in the series expansion of $f(z)$ are equal to $1$, and $\Lambda$ is not reciprocal summable.
A: Recall  Pólya-Carlson's theorem${}^{\color{blue}{[1]}}$

If $$f(z) = \sum_{n=0}^\infty c_n z^n$$ has integer coefficients and
  is analytic for $|z| < 1$, then either $|z| = 1$ is a 
  natural boundary
  for $f$, or $f$ is rational.

Let's consider what happens when $f$ is rational. Since $\Lambda$ is infinite,
$f$ cannot be a polynomial. Express $f(x)$ as a ratio of two polynomials
$$f(x) = \frac{N(x)}{D(x)} \quad\iff\quad D(x)f(x) = N(x)$$
and let $D(x) = 1 - d_1 x - \cdots - d_m x^m$. For $n > \max(m,\deg N)$,
$c_n$ satisfies a linear recurrence relation:
$$c_n = d_1 c_{n-1} + d_2 c_{n-2} + \cdots + d_m c_{n-m}$$
Consider following mapping $\varphi : \{0,1\}^m \to \mathbb{R}^m$ which sends
$$(x_0,x_1,\ldots,x_{m-1})\quad\mapsto\quad (x_1,\ldots, d_1 x_{m-1} + d_2 x_{m-2} + \cdots + d_m x_0)$$
Let $(e_n)$ be the sequence of finite subsequences of $(c_n)$ of length $m$.
i.e. 
$$e_n = (c_n, c_{n+1}, \ldots, c_{n+m-1})$$
When $n > \deg N$, $\varphi$ sends $e_n$ to $e_{n+1} \in \{0,1\}^m$.
Since the state space of $(e_n)$, $\{0,1\}^m$, is finite, the sequence $(e_n)$ and hence the sequence $(c_n)$ will be ultimately periodic. In this case, $\lambda_n$ cannot be reciprocal summable.
This implies Claim I is true. 
Notes


*

*$\color{blue}{[1]}$ - see $\S 6.5$ of Sanford L. Segal's Nine introductions in Complex Analysis.

