<s>[INST] Calculus - Infinite Series Decide whether the infinite series converges of diverges.
$$\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$$
My thought process: 
   The nth term test doesn't seem viable after initial use. 
   The comparison function I derived for DCT/LCT is: $\frac{3^n}{4^n}$.
   I cannot create an appropriate inequality between the given function and my comparison function I found, so I tried LCT. I am aware that 
$\sum_{i=1}^\infty \frac{3^n}{4^n}$ is geometric and because the common ratio (r) is $ \frac{3}{4}$, $-1<r<1$ the series converges, however I cannot compute the limit LCT requires. 
Application of the root test would seem unbeneficial because roots do not split over sums. 
As for the ratio test I was also unsuccessful in proceeding with the computation of the limit. 
The above are the only tests I have learned as of right now. My question is what test should I be looking to utilize to determine convergence or divergence? 
 [/INST] Besides the already provided solutions, I'd like to point out that both the LCT and the Root Test do work here too.
If we're applying the LCT to $\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$ and $\sum_{n=1}^\infty \frac{3^n}{4^n}=\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$:
$$\frac{\frac{2^n+3^n}{3^n+4^n}}{\frac{3^n}{4^n}}=\frac{2^n+3^n}{3^n+4^n}\cdot\frac{4^n}{3^n}=\frac{8^n+12^n}{9^n+12^n}=\frac{(8^n+12^n)\color{red}{\div12^n}}{(9^n+12^n)\color{red}{\div12^n}}=\frac{\left(\frac{8}{12}\right)^n+1}{\left(\frac{9}{12}\right)^n+1}\to\frac{0+1}{0+1}=1,$$
so the LCT applies to these two series and they behave the same way.
A similar trick works with the Root Test too. We can factor our $3^n$ and $4^n$ in the numerator and the denominator, respectively, to find that
$$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}=\sqrt[n]{\frac{\color{red}{3^n\cdot}\left(\left(\frac{2}{3}\right)^n+1\right)}{\color{red}{4^n\cdot}\left(\left(\frac{3}{4}\right)^n+1\right)}}=\color{red}{\frac{3}{4}\cdot}\sqrt[n]{\frac{\left(\frac{2}{3}\right)^n+1}{\left(\frac{3}{4}\right)^n+1}}\to\frac{3}{4}\cdot1=\frac{3}{4}.$$
UPDATE: I've just noticed that the OP mentioned the Ratio Test too. I'm not going to type one more solution, but the limit of the fraction in the Ratio Test can be simplified here using a trick similar to the one used in the LCT calculation above.
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