<s>[INST] 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.
 [/INST] 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, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^2+1} \\  \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^3+n^3} \\  = \sum_{n=1}^\infty \frac{n^2-7n}{5n^3} \\ = \sum_{n=1}^\infty \frac{1}{5n}-\sum_{n=1}^\infty\frac{7}{5n^2} \\ = \frac{1}{5}\sum_{n=1}^\infty \frac{1}{n}-\frac{7\pi^2}{30}$$ 
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