Q:

Test the series for convergence

A pretty nasty one in my opinion, due to the logarithm. I'm stuck and don't know what test to use. Some help would be appreciated.
$$\sum_{n=1}^{\infty} \ln(3n)\Bigl(n\sin\frac{1}{n}\Bigr)^{\!n^3}$$

A:

Since you have 
$$
\sin x = x-\frac{x^3}{6} + o(x^4)\tag{1}
$$
when $x\to 0$, we can write, when $n\to \infty$,
$$
n\sin\frac{1}{n} = n\left(\frac{1}{n}-\frac{1}{6n^3} + o\left(\frac{1}{n^4}\right)\right) = 1-\frac{1}{6n^2} + o\left(\frac{1}{n^3}\right)\tag{2}
$$
Plugging this into $(n\sin\frac{1}{n})^{n^3}$, we get
$$
\left(n\sin\frac{1}{n}\right)^{n^3}
= \left(1-\frac{1}{6n^2} + o\left(\frac{1}{n^3}\right)\right)^{n^3}
= \exp\left(n^3\ln\left(1-\frac{1}{6n^2} + o\left(\frac{1}{n^3}\right)\right)\right)
$$
and using the Taylor series expansion of $\ln(1-x)= -x+o(x)$ when $x\to0$, we get finally
$$
\left(n\sin\frac{1}{n}\right)^{n^3}
= \exp\left(-\frac{n}{6} + o\left(1\right)\right)
\operatorname*{\sim}_{n\to\infty} e^{-\frac{n}{6}}\,.\tag{3}
$$
Therefore, overall, since $\ln(3n) \operatorname*{\sim}_{n\to\infty} \ln n$, we have
$$
\ln(3n)\left(n\sin\frac{1}{n}\right)^{n^3}
\operatorname*{\sim}_{n\to\infty} (\ln n) e^{-\frac{n}{6}}
$$
and the series converges (very fast) by comparison with $\sum_{n} (\ln n) e^{-\frac{n}{6}}$.

