I'm looking for tips on how to solve this infinite series problem.
$\sum_{x=0}^{\infty}\frac{x}{4^x}$
I am pretty sure this series converges, but I don't know which method I should use to tackle this problem. At first I thought it was geometric, but the ratio isn't a constant: $r = \frac{a_{n+1}}{a_{n}} = \frac{x+1}{4x}$.
When this happens, how should I proceed?
Let $f(t)=\sum\limits_{x=0}^{\infty} t^{x}$. Then $tf'(t)=\sum\limits_{x=0}^{\infty}x t^{x}$. Put $t=\frac 1 4$. Now note that $f(t)=\frac 1 {1-t}$ so we can compute $tf'(t)$ when $t=\frac 1 4$.
In addition to using calculus, as shown in an answer already, here's an alternative approach which doesn't require it. (This is also really similar to how normal geometric sums and series are evaluated.) You can evaluate the series without it by shifting all the powers of $4$ as follows:
$$S_n = \sum_{x = 0}^{n}\frac{x}{4^x} = \frac{1}{4}\color{blue}{+\frac{2}{16}+\frac{3}{64}+\dots+\frac{n}{4^n}} \tag{1}$$
$$\frac{1}{4}S_n = \frac{1}{4}\sum_{x = 0}^{n}\frac{x}{4^x} = \color{blue}{\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\dots+\frac{n-1}{4^{n+1}}}+\frac{n}{4^{n+1}} \tag{2}$$
For $(1)-(2)$, notice how shifting all the $x$'s to one higher power of $4$ allows the blue terms to form a geometric sum: $\color{blue}{\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dots+\dfrac{1}{4^{n+1}}}$.
So, $(1)-(2)$ gives:
$$\frac{3}{4}S_n = \frac{1}{4}+\color{blue}{\left(\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dots+\dfrac{1}{4^{n+1}}\right)}-\frac{n}{4^{n+1}}$$
$$S_n = \frac{1}{3}+\frac{4}{3}\left(\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dots+\dfrac{1}{4^{n+1}}\right) -\frac{n}{3\cdot 4^n}$$
From here, you want $S_\infty$, which is just $\lim_\limits{n \to \infty} S_n$. Given that you already know how to deal with convergent geometric series, it should be pretty simple. Also, keep in mind that exponentials outgrow polynomials, so the last term goes to $0$ as $n \to \infty$.
We can use the inequality $x\le 2^x$ with $x\in\{0,1,\ldots\}$ to establish the convergence and obtain an upper bound in the following way $$ \sum_{x=0}^\infty\frac{x}{4^x} \le\sum_{x=0}^\infty\frac{2^x}{4^x} =\sum_{x=0}^\infty\Bigl(\frac12\Bigr)^x=2. $$
