Finding convergence using Limit Comparison Test What should I set as my second function to determine the convergence of
$$ \sum_{n=1}^\infty \frac{\ln(n)} {\mathbb e^n\sqrt n} $$
by the limit comparison test? I'm not too sure where to begin, any help is appreciated.
 A: From this plot using Wolfram Alpha, here, it can be said that,
$$\frac{ln(n)}{\sqrt{n}e^n} \leq \frac{1}{n^2} \forall n \in [1,\infty)$$
Hence using the limit comparison test it can be said that,
$ \sum_{n=1}^\infty \frac{ln(n)} {(\sqrt[2]n)e^n} $ is convergent because,
$ \sum_{n=1}^\infty \frac{1} {n^2} $ is convergent.
$$\lim_{n\to\infty}a_n/b_n = \lim_{n\to\infty} \frac{ln(n)/\sqrt{n}e^n}{1/n^2}$$
$$\implies \lim_{n\to\infty} \frac{ln(n)n^{3/2}}{e^n}$$
Since the above limit is in the $\infty/\infty$ form, we differentiate using the L'Hospital Rule.
$$\implies \lim_{n\to\infty} \frac{\frac{3}{2}n^{1/2}ln(n) + n^{1/2}}{e^n}$$
$$\implies \lim_{n\to\infty} \frac{\frac{3}{4}n^{-1/2}ln(n) + \frac{3}{2}n^{-1/2}ln(n) + \frac{1}{2}n^{-1/2}}{e^n}$$
$$\implies \lim_{n\to\infty} \frac{\frac{9}{4}n^{-1/2}ln(n)}{e^n}$$
$$\implies \lim_{n\to\infty} \frac{\frac{9}{4}ln(n)}{\sqrt{n}e^n}$$
$$\implies \lim_{n\to\infty} \frac{\frac{9}{4n}}{\sqrt{n}e^n + e^n(\frac{1}{2\sqrt{n}})}$$
$$\implies \lim_{n\to\infty} \frac{\frac{9.2\sqrt{n}}{4n}}{e^n(2n+1)}$$
$$\implies \lim_{n\to\infty} \frac{18\sqrt{n}}{e^n(8n^2+4n)}$$
$$\implies \lim_{n\to\infty} \frac{9/\sqrt{n}}{e^n(16n+4) + e^n(8n^2+4n)} = 0$$
A: $$
\frac{\ln n}{e^n\sqrt n} \le \frac {\sqrt n}{ e^n\sqrt n} = \frac 1 {e^n},
$$
so the series converges.
