Limit of ratio between a power series and a "subset" of the power series $B$ is an infinite power series that converges everywhere, and $A$ is an infinite power series that converges everywhere which is composed only of terms found in $B$ - both have nonnegative real coefficients and are series in a single, nonnegative real variable, $x$, thus both are strictly increasing in $x$. 
Are there conditions on $A$ or $B$ which determine that $A/B$ has a limit, (i.e. it doesn't oscillate) as $x \to \infty$? It certainly doesn't diverge because $B\ge A$.
For my problem, I have a strong feeling, based on some heuristic reasoning and based on strong evidence from a graph, that $A/B$ converges to some limit, but I am having the darnest time proving it.
Any help greatly appreciated!
EDIT:
The actual series' I have in mind are:
$$ B = e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!} $$
and, for fixed $n\in\mathbb{Z^+}$
$$ A = \sum_{j=0}^{\infty} \frac{x^{jn}}{(jn)!} $$
In other words, $A$ is composed of every $n$th term of $e^x$, starting with the $0$th term.
 A: In the case you gave the answer can be obtained through some formal manipulations.
Claim.
$$
\sum_{j=0}^{\infty} \frac{x^{jn}}{(jn)!} = \frac{1}{n} \sum_{k=0}^{n-1} \exp\left(xe^{i2\pi k/n}\right).
$$
Proof.
Because the series in question are absolutely convergent we may rearrange the terms as follows:
$$
\begin{align}
\sum_{k=0}^{n-1} \exp\left(xe^{i2\pi k/n}\right) &= \sum_{k=0}^{n-1} \sum_{j=0}^{\infty} \frac{x^j e^{i2\pi kj/n}}{j!} \\
&= \sum_{j=0}^{\infty} \frac{x^j}{j!} \sum_{k=0}^{n-1} e^{i2\pi kj/n}.
\end{align}
$$
When $j$ is a multiple of $n$ all of the terms of the inner sum are equal to $1$ so the sum is equal to $n$.  When $j$ isn't a multiple of $n$ we note that the inner sum is the partial sum of a geometric series, so
$$
\sum_{k=0}^{n-1} e^{i2\pi kj/n} = \frac{1 - e^{i2\pi j}}{1 - e^{i2\pi j/n}} = 0.
$$
Consequently the only terms remaining in the sum are those where $j$ is a multiple of $n$.  We thus have
$$
\sum_{k=0}^{n-1} \exp\left(xe^{i2\pi k/n}\right) = n \sum_{j=0}^{\infty} \frac{x^{jn}}{(jn)!}.
$$
Q.E.D.
Using this formula we find that
$$
\frac{\displaystyle\sum_{j=0}^{\infty} \frac{x^{jn}}{(jn)!}}{\displaystyle\sum_{j=0}^{\infty} \frac{x^{j}}{j!}} = \frac{1}{n} + \frac{1}{n} \sum_{k=1}^{n-1} \exp\left(x\left(e^{i2\pi k/n}-1\right)\right) \longrightarrow \frac{1}{n}
$$
as $x \to \infty$.
