Upper bound the Polylogarithm $\sum_{n=1}^\infty \frac{x^n}{n^2}$

Question

Let $x \in (0,1)$ be some real number, we can then consider the Polylogarithm: $$\operatorname{L}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$$ It is not hard to see that the following upper bound holds: $$ \operatorname{L}_2(x) = x \cdot \sum_{n=1}^\infty \frac{x^{n-1}}{n^2} \leq x \sum_{n=1}^\infty \frac{1}{n^2} =\frac{x\pi^2}{6}. $$ One could continue to get a finer upper bound using the same strategy: $$ \operatorname{L}_2(x) = \sum_{n=1}^m \frac{x^n}{n^2} + \sum_{n=m+1}^\infty \frac{x^{n}}{n^2} $$ where we can then upper bound $$\sum_{n=m+1}^\infty \frac{x^{n}}{n^2} \leq x^{m+1} \cdot \sum_{n=m+1}^\infty \frac{1}{n^2}$$ I am quite pleased with this, but was wondering if there are alternative, better well known bounds.