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Here's a series the limit of which I want to evaluate:

$$a_n=\sum_{n=1}^{\infty}\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1}$$

Here's how I did it. Let's define 3 sequences:

$b_n=\frac{n^2-7n}{n^3+3n^2+n}$, $c_n=\frac{n-\frac{n}{2}}{n^2+4n}$, $d_n=\frac{2}{3n}$

We have $\sum d_n=\frac{2}{3}\sum\frac{1}{n}$ and as $\sum\frac{1}{n}$ is divergent, $\sum d_n$ is also divergent.

So if we prove that for sufficiently large n $d_n \leq c_n$, same for $c_n \leq b_n$ and finally same for $b_n \leq a_n$ then by comparison test we will prove, that $\sum a_n$ is divergent. So let's do this now.

$\sum c_n=\sum\frac{1}{2(n+4)}=\frac{1}{2}\sum\frac{1}{n+4}$

Now let $e_n=\frac{1}{n+4}$. For sufficiently large n we have $\frac{1}{n+\frac{n}{2}}=\frac{2}{3n}=d_n\leq e_n$. So, by comparison test $\sum e_n$ divergent, so that means $\sum c_n$ is also divergent.

$c_n \leq b_n$ because $\frac{n-\frac{n}{2}}{n^2+3n+n} \leq \frac{n-7}{n^2+3n+1}=\frac{n^2-7n}{n^3+3n^2+n}$and $b_n \leq a_n$ is obvious.

So this is it, now we've proven that $\sum a_n$ is divergent. The problem is it took a while and I'm not sure whether my solution is correct. Are there other, simpler ways to solve this problem? Preferably without derivatives and integrals - I haven't covered them in my course yet.

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Use the limit comparison test. Let $a_n = \dfrac 1n$ and $$b_n = \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1}.$$

Since $\displaystyle \sum_{n=1}^\infty a_n$ diverges and $\dfrac{a_n}{b_n} \to \dfrac 1e < \infty$, you have that $\displaystyle \sum_{n=1}^\infty b_n$ diverges too.

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You can use a simpler approach.

$a_n=\sum_{n=1}^{\infty}\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1}=\sum_{n=1}^{\infty}\frac{(1+\frac{1}{n})^nn^2}{n^3+3n^2+1}-7\sum_{n=1}^{\infty}\frac{n}{n^3+3n^2+1}$

Ok, let's call $b_n=\sum_{n=1}^{\infty}\frac{(1+\frac{1}{n})^nn^2}{n^3+3n^2+1}$ and $c_n=\sum_{n=1}^{\infty}\frac{n}{n^3+3n^2+1}$

$c_n$ converges easily...

And $b_n>\sum_{n=1}^{\infty}\frac{n^2}{n^3+3n^2+1}$, which diverges...

So $a_n$ diverges.

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