Check convergence of $\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$ Zoomed version: $$\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$$
So, I've seen similar example at Convergence or divergence of $\sum \frac{3^n + n^2}{2^n + n^3}$
And I liked that answer : 

$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\infty2^n,\enspace\text{hence}\quad \frac{3^n+n^2}{2^n+n^3}\sim_\infty\Bigr(\frac32\Bigl)^n,$$
  which doesn't even tend to $0$.

Could you explain why does $3^n+n^2\sim_03^n$ and $2^n+n^3\sim_\infty2^n$ when we have only $n \to \infty$ ?

The notation $$3^n+n^2\sim_{+\infty}3^n$$ means $$\lim_{n \to +\infty} \frac{3^n+n^2}{3^n}=1.$$ You can see this is true since $$\lim_{n \to +\infty} \frac{3^n+n^2}{3^n} = 1 + \lim_{n\to +\infty} \frac{n^2}{3^n}.$$ The second limit is well-known to be equal to $0$.
