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 $0$.
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.
An intuitive approach
We wish to compare $3^n$ and $n^2$ for $n\in \Bbb{N}$
Let's consider the sequence $(a_n,b_n)=(3^n,n^2)$ and assign some values to $n$. We get $$n=1\rightarrow (a_1,b_1)=(3,1)|a_1=3b_1\\n=2\rightarrow (a_2,b_2)=(9,4)| a_2\gt2b_2\\n=3\rightarrow (a_3,b_3)=(27,9)|a_3\gt3b_3\\n=4\rightarrow (a_4,b_4)=(81,16)|a_4\gt5b_4\\n=5\rightarrow (a_5,b_5)=(243,25)|a_5 \sim10b_5\\n=6\rightarrow (a_6,b_6)=(729,36)|a_6\gt20b_6\\\cdots$$
A geometric/intuitive approach
Consider the green vertical line as any value of n you wish to assign (here it is $n=5$) and extrapolate intuitively to $\infty$.
A more rigorous approach
For a fixed $k\in \Bbb{N}$ we have $$\lim_{n\rightarrow \infty}3^n\gt\lim_{n\rightarrow \infty}n^k\gt\lim_{n\rightarrow \infty}n^{k-1}\gt\cdots \gt\lim_{n\rightarrow \infty}n^3\gt\lim_{n\rightarrow \infty}n^2$$
So, for as large a $k$ as you wish, $3^n$ will be larger than $n^k$ as $n\rightarrow \infty$ and as you can easily see $n^k$ being as larger than $x^3$ as you wish, you get an even larger value for $3^n$.
