I need to conclude if the following series is convergent
$$ \sum_{i=1}^\infty \frac{3^n+n^2}{2^n+n^3}. $$
Can I get a hint? I tried to calculate $\dfrac{a_{n+1}}{a_{n}}$ and to see if the series is monotonically increasing and therefore divergent, but it seems like a difficult way.
Thanks!
$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\infty2^n,\enspace\text{hence}\quad \frac{3^n+n^2}{2^n+n^3}\sim_\infty\Bigr(\frac32\Bigl)^n,$$ which doesn't even tend to $0$.
Apply the divergence test:
$$ \lim_{n \to \infty} \frac{3^n+n^2}{2^n+n^3} \stackrel{4 \times \text{L'Hôpital}}{=} \frac{\ln^4 3}{\ln^4 2} \lim_{n \to \infty} \left(\frac{3}{2}\right)^n $$
As the series tends to infinity, it behaves like a geometric series with $|r| = \frac{3}{2} > 1$.
$$\frac 1n<\frac 1n\left(\frac{n3^n+n^3}{2^n+n^3}\right)=\left(\frac{3^n+n^2}{2^n+n^3}\right)$$ Hence the series is divergent because the harmonic series $\sum \frac 1n$ diverges.
