What you should know is the fact that if $Y$ and $Z$ are independent random variables, then the distribution function of $Y+Z$ is the convolution of the distribution functions of $Y$ and $Z$. So the convolution of two independent random variables(or distributions) represents their sum.
With that in mind, the sum $\sum_{n=1}^\infty F^{n}(x) = \sum_{n=1}^\infty P(X_1+...+X_n \leq x)$ where $X_1,...,X_n$ are iid with distribution $F$.
To show that this sum is finite, we basically need to focus on large $n$, and showing that for large $n$, the terms are very small. How? Well, the $X_i$ are non-negative integer valued : so if $n$ is an integer much larger than $x$, then for $X_1+...X_n \leq x$ to happen, a lot of the $X_i$ will have to be zero. The condition $F(0)<1$ guarantees that this can only happen with a certain probability, and by independence we will get a bound that should work.
More precisely : if $X_1+...+X_n \leq x$ for some $n > \lceil x\rceil$, then at least $n-\lceil x \rceil$ of the $X_i$ are zero. So, we bound :
$$
P(X_1+...+x_n \leq x) \leq P(\text{at least $n-\lceil x\rceil$ of the $X_i$ are zero}) \\
\leq \sum_{j=n-\lceil x\rceil}^n P(\text{at least $j$ of the $X_i$ are zero})\\ \leq\sum_{j=n - \lceil x \rceil}^n\binom{n}{j} F(0)^{n-j} (1-F(0))^{j} \\
= \sum_{j=0}^{\lceil x \rceil} \binom{n}{j} F(0)^j (1-F(0))^{n-j}
$$
This is the probability that $Bin(n,F(0)) \leq \lceil x \rceil$, where $Bin(n,p)$ is a binomial random variable. Look up tail bounds for the binomial random variable like Hoeffding etc. and see if you can finish this off by yourself.
From the gaussian assumption, we can calculate analytically the two probabilities, it suffices to notice that $\sum_{i=1}^nX_i\sim \sqrt{n}\mathcal{N}(0,1)$ then
$$\mathbb{P}\left(\sum_{i=1}^nX_i\le c \right) = \mathbb{P}\left(\mathcal{N}(0,1)\le \frac{c}{\sqrt n} \right) = \Phi\left(\frac{c}{\sqrt n}\right)$$
As $\frac{c}{\sqrt n} <\frac{c}{\sqrt m}$ for $c \ge 0$ and $n >m$, we have
$$\mathbb{P}\left(\sum_{i=1}^nX_i\le c \right)\le \mathbb{P}\left(\sum_{i=1}^mX_i\le c \right)$$
Best Answer
Are you sure this is the right question? Suppose $X_i \overset{i.i.d.}{\sim} Bernoulli(1/2)$. Then $F(0) = 1/2 < 1$ and $F(1) = 1$, so no it doesn't follow.