Does $\sum\ln(n) / n^3$ converge? By Direct Comparison Test:
$$\ln(n)/n^3  < n/n^3  = 1/n^2  $$
Converges as that is a convergent a p-series
By n'th term Test:
$$(\ln(n))' = 1/n$$
$$(n^3)' = 3n^2$$
$$\lim_{n\to\infty} (1/n) / 3n^2 = \lim_{n\to\infty} (3n^2)/n =\lim_{n\to\infty} 3n  = \infty$$
so by $n$'th term it diverges.
What am I doing wrong?
 A: You made an algebraic mistake. You have the quotient $$ 1/n\over 3n^2.$$  If you wish to "move the numerator $1/n$ downstairs", you need to take its reciprocal, ${1\over 1/n}=n$, first and do it:
$$
{\color{maroon}{1/n}\over 3n^2}={1\over \color{maroon}n\cdot 3n^2}.
$$
You did this correctly. 
However, when you "moved the denominator $3n^2$ upstairs", you failed to use its reciprocal $1/3n^2$. Done correctly, you would have obtained
$$
{1/n\over\color{maroon}{ 3n^2}}={(1/n)\cdot(\color{maroon}{1/3n^2})\over1 }.
$$
Note both methods give $1\over 3n^3$ as a result; there is no need to do both. Perhaps the simplest thing to do in order to simplify your expression is to write
$$
{1/n\over 3n^2}={1\over n}\cdot{1\over 3n^2}={1\over n\cdot3n^2}={1\over 3n^3}.
$$
A: You have a "typo". The terms do go to zero: $\displaystyle \lim_{n\to \infty}\frac{(1/n)}{3n^2}=\lim_{n\to\infty}\frac{1}{3n^3}=0$. 
A: You got the division wrong. 
$$(1/n)/3n^2 = 1/3n^3 \rightarrow 0$$
A: You're nth term test is wrong. The reason why you're getting a divergent limit is because you flipped your fractions around inccorectlly. Anyways, the nth term test dosen't involve taking derivitives. But, here is how you can do it:
$\frac{ln(n)}{n^3}>0$ for sufficently large n (i.e n>3). Now, we have that $0<\frac{ln(n)}{n^3}<\frac{1}{n^2}$. Clearly, $\frac{1}{n^2}\rightarrow 0$. Hence, by the squeeze theorem for limits, $\frac{ln(n)}{n^3}$ $\rightarrow 0$. Also, always remember that the nth term test is not sufficent to prove convergance, it's just a nessacary condition. i.e the harmonic series.
