We will use the statement
"If $f(x) \to L$ as $x \to \infty$ for real $x$, then $f(n) \to L$ as $n \to \infty$ for natural $n$."
This is so we can use L'Hôpital's Rule since it does not work for functions with natural inputs. Let us also use your idea of leveraging exponential and logarithm properties. Also, we will omit using absolute values because the function is always non-negative.
Let $f(x) = \left(\dfrac{2^{x}}{3^{x}+4^{x}}\right)^{\frac{1}{x}}$. As $x \to \infty$, we evaluate the limit as follows:
$$
\eqalign{
\displaystyle
\lim_{x\to\infty} \left(\frac{2^{x}}{3^{x}+4^{x}}\right)^{\frac{1}{x}} &= \lim_{x\to\infty} \exp\left(\ln\left(\left(\frac{2^{x}}{3^{x}+4^{x}}\right)^{\frac{1}{x}}\right)\right) \cr
&= \lim_{x\to\infty} \exp\left(\frac{\ln\left(\frac{2^{x}}{3^{x}+4^{x}}\right)}{x}\right) \cr
&= \exp\left(\lim_{x\to\infty}\frac{\ln\left(\frac{2^{x}}{3^{x}+4^{x}}\right)}{x}\right) \cr
&= \exp\left(\lim_{x\to\infty} \frac{\frac{d}{dx}\ln\left(\frac{2^{x}}{3^{x}+4^{x}}\right)}{\frac{d}{dx}x}\right) \cr
&= \exp\left(\lim_{x\to\infty}\frac{2^{2x}\left(\ln\left(2\right)-\ln\left(4\right)\right)+3^{x}\left(\ln\left(2\right)-\ln\left(3\right)\right)}{2^{2x}+3^{x}}\right) \cr
&= \exp\left(\lim_{x\to\infty}\frac{\left(\ln\left(2\right)-\ln\left(4\right)\right)+3^{x}2^{-2x}\left(\ln\left(2\right)-\ln\left(3\right)\right)}{1+3^{x}2^{-2x}}\right) \cr
&= \exp\left(\ln\left(2\right)-\ln\left(4\right)\right) \cr
&= \dfrac{1}{2}.
}
$$
Therefore,
$$\lim_{n\to\infty} \left(\frac{2^{n}}{3^{n}+4^{n}}\right)^{\frac{1}{n}} = \dfrac{1}{2}.$$
By the Root Test, we conclude that the series in question converges absolutely (hence converges).
I tried leveraging logarithm properties and rewriting the denominator as $e^{\ln{((3^n+5^n)^\frac{1}{n})}}$, or $e^\frac{\ln{(3^n+5^n)}}{n}$, and then l'Hopital-ing the fraction, but I got stuck in an infinite derivation loop.
You cannot use L'Hôpital's Rule for functions with integer inputs. The rule requires continuous functions to use derivatives.
Please let me know if you have any questions.