From the continuous fraction expansion, the seventh convergent is
$$\gamma \approx \frac{15}{26}$$
From the limit definition
$$\begin{align}
\gamma
&=
\lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\
&=
\frac{7}{12}+\sum_{n=1}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
&=\frac{7}{12}-\frac{1}{180}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
&=\frac{26}{45}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
\end{align}
$$
so
$$\gamma \approx \frac{26}{45}$$
Multiplying both approximations,
$$\gamma^2 \approx \frac{1}{3}$$
The origin of this limit definition is improving the convergence of Macys formula from $o(n^{-2})$ to $o(n^{-4})$ downweighting the last term, without additional fractions (https://math.stackexchange.com/a/129808/134791)
In fact, the convergent approximation is not necessary.
Given $$\gamma \approx \frac{26}{45}$$
we have
$$3\gamma^2\approx3\left(\frac{26}{45}\right)^2=3\frac{676}{2025}=3\frac{676}{3\cdot675}=\frac{676}{675}\approx 1$$
which also yields
$$\gamma^2 \approx \frac{1}{3}$$
Towards proving that $\gamma < \frac{1}{ \sqrt {3} }$, we may take one more term out of the summation:
$$\gamma \approx \frac{7}{12}-\frac{1}{180}-\frac{1}{2310}=\frac{4001}{6930}<\frac{1}{\sqrt{3}}$$
Although your method is correct and works, the method to determine convergence that seems most obvious to me (and the easiest to do) is the direct comparison test with $\frac{1}{n}$. Specifically $$\frac{\ln(n)}{n} > \frac{1}{n}$$ for $n \ge 3$. Therefore $$\sum_{n=1}^\infty \frac{\ln(n)}{n} > \sum_{n=1}^\infty \frac{1}{n}$$ which is known to diverge.
Best Answer
It is unknown whether $$ \sum_{n=1}^\infty\frac{1}{n^3\sin^2n} $$ converges or not. The difficulty here is that convergence depends on the term $n\sin n$ not being too small, which in turn depends on how well $\pi$ can be approximated by rational numbers. It is possible that, if $\pi$ can be approximated `too well' by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is $$ x_n=\frac{1}{n^2\sin n}. $$ We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $\vert p/q-\pi\vert\le q^{-3+\epsilon}$ (for any $\epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-\epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $\vert p/q-\pi\vert\le q^{-3-\epsilon}$, then infinitely many $x_n$ would be of order at least $n^\epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $\pi$. The sequence $x_n$ converges to zero if the irrationality measure of $\pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$* (see the link to the mathworld page above). It is expected that the irrationality measure of $\pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
[* The best known bound for the irrationality measure, as listed on the mathworld page, has been improved! It is now 7.10320533, according to the (not yet peer-reviewed) paper by Zeilberger and Zudlin, 2019. This is still much too small an improvement to say whether or not the sequence above converges.]