Did I correctly verify the convergence of this series? I want to find if the following series is convergent.
$$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I use the asymptotic criterion for series convergence.
$$ a_n=\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} $$
I take such $b_n$ that $a_n$ and $b_n$ are asymptotically similar and that the convergence of $\sum_{n=1}^\infty b_n$ is known.
$$b_n=\frac{1}{n}$$
$$\lim_{n\to \infty}\frac{a_n}{b_n}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \frac {n}{1}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^3-7n^2}{n^3+3n^2+1}$$
The limit is $e$ which proves that $a_n \sim b_n$.
Then since $\sum_{n=1}^\infty \frac{1}{n} $ is divergent, so is the original series.
I'd be thankful if someone could review this and tell me if this solution is correct.

Your conclusion is correct. You could get there using the comparison test too if you are interested. Namely, that $$1 < \left(1+\frac{1}{n} \right)^n   \\ \frac{1}{n^3+3n^3+n^3}\leq \frac{1}{n^3+3n^2+1}$$ for all $n \geq 1$. Hence, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^2+1} \\  \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^3+n^3} \\  = \sum_{n=1}^\infty \frac{n^2-7n}{5n^3} \\ = \sum_{n=1}^\infty \frac{1}{5n}-\sum_{n=1}^\infty\frac{7}{5n^2} \\ = \frac{1}{5}\sum_{n=1}^\infty \frac{1}{n}-\frac{7\pi^2}{30}$$ 
