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 $n \to \infty$ ?

To begin, notice that for all $n\geq 1$ one has $2^n\geq n^2$
$$\sum\limits_{n=1}^\infty \frac{2^n+n^2}{3^n+n^3}\leq \sum\limits_{n=1}^\infty\frac{2^n+2^n}{3^n+n^3}\leq 2\sum\limits_{n=1}^\infty \frac{2^n}{3^n}= 4$$
Since all terms of the original series are non-negative and the series is bounded above, it must converge by the dominated convergence theorem.
Note: the first inequality is true since $n^2\leq 2^n$.
The second inequality is true since $n^3\geq 0$, we are dividing by a smaller denominator at each step thereby making each number larger.
The final equality is true by properties of geometric series.

A more relaxed approach is to simply note that $2^n+n^2$ "acts like" $2^n$, and similarly $3^n+n^3$ "acts like" $3^n$, to note that $\sum\limits_{n=1}^\infty \frac{2^n+n^2}{3^n+n^3}$ "acts like" $\sum\limits_{n=1}^\infty \frac{2^n}{3^n}$
To use this in action, we may use instead of direct comparison as above, the ratio test.
If $\lim\limits_{n\to\infty}|\frac{a_{n+1}}{a_{n}}|<1$ then $\sum\limits_{n=1}^\infty a_n$ converges.
Here, we look at $\frac{a_{n+1}}{a_n} = \frac{2^{n+1}+(n+1)^2}{3^{n+1}+(n+1)^3}\cdot \frac{3^n+n^3}{2^n+n^2} = \frac{2\cdot 6^n + 2^{n+1}n^3+3^n(n+1)^2+n^3(n+1)^2}{3\cdot 6^n + 2^n(n+1)^3+3^{n+1}n^2+(n+1)^3n^2}$
By dividing top and bottom by $6^n$ and taking the limit as $n\to\infty$ we see that $\frac{a_{n+1}}{a_n}\to \frac{2}{3}$ since every term but the first for each of the numerator and denominator approach zero.
