Finding the limit of a sequence involving alternating signs. Determine whether the following sequence converges and if so, find its limit.
$$\bigg(\frac{(-1)^ne^n}{\ln^2(n)}\bigg)^\infty_{n=2}$$
I can't figure out how to proceed with this question. Is it perhaps L'Hopital? But then how would you calculate the derivative of the numerator? I'm thinking it's most likely to be the Squeeze Theorem because of that $(-1)^n$. Any help would be most appreciated.
 A: Observe that
$$
\ln n\leq n^{1/4}\implies \ln^2n\leq n^{1/2}
$$
for sufficiently large $n$. Also $e^n\ge 1+n$. Thus
$$
\left\lvert
\frac{(-1)^ne^n}{\ln^2(n)}
\right\rvert
=\left\lvert
\frac{e^n}{\ln^2(n)}
\right\rvert
\ge\frac{1+n}{n^{1/2}}
$$
Let $n\to\infty$ and remember that a necessary condition for a sequence to converge is  that it be bounded.
A: The absolute value of the $n$th term is $\dfrac{e^n}{\ln^2 n} \to \infty.$ How could it converge?
A: Using Cauchy's Criterion, you can see that the series diverges, because:
$$
\sqrt[n]{\left\lvert\frac{(-1)^ne^n}{\ln^2(n)}\right\rvert}
=\frac{e}{\ln^{\frac{2}{n}(n)}}
\to e>1
$$
A: Here's my own answer.
For $n \geq e$, we have $e^n \geq n$. Therefore, 
$$\frac{e^n}{\ln^2(n)}\geq\frac{n}{\ln^2(n)}$$
By L'Hopital's Rule ($\frac{\infty}{\infty}$ form), we have:
$$\lim_{n\to\infty}\frac{n}{\ln^2(n)}=\lim_{n\to\infty}\frac{1}{\frac{2\ln(n)}{n}}=\lim_{n\to\infty}\frac{n}{2\ln(n)}$$
Through the second use of L'Hopital's Rule;
$$\lim_{n\to\infty}\frac{n}{2}=\infty$$
Hence, the sequence $\big(\frac{e^n}{\ln^2(n)}\big)^{\infty}_{n=1}$ is unbounded and so $\bigg((-1)^n\frac{e^n}{\ln^2(n)}\bigg)^{\infty}_{n=1}$ diverges.
