Determine the end behavior of $\sum_{n=0}^{\infty}\frac{2^n}{3^n+4^n}$ using Root Test I took the nth root of $\frac{2^n}{3^n+4^n}$, and while the numerator simplifies nicely to 2, the denominator is totally stumping me. Since this would all be inside of an infinite limit, we must consider the limit of $(3^n+4^n)^\frac{1}{n}$ as n goes to $\infty$, but we end up with a indeterminate situation of $\infty^0$.
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.
I'm specifically looking for a Root Test solution.
 A: Hint: $3^n + 4^n > 4^n$, so $(3^n + 4^n)^{1/n} > 4$.
A: 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.
