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In this answer it is stated that $$ \sum_{n\geq1}\frac{1}{n^2}=\sum_{n\leq x}\frac1{n^2}+\mathcal O(1/x). $$ Is this statement true as $x\to\infty$?

What I've done: If $x$ is fixed, then I think the answer is almost trivial, because we may set $C=\pi^2x/6$, so $$ \sum_{n=x}^\infty\frac1{n^2}\leq\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}=\frac{C}{x}, $$ therefore $$ \sum_{n\geq1}\frac1{n^2}=\sum_{n\leq x}\frac{1}{n^2}+\sum_{n=x}^\infty\frac{1}{n^2}\leq\sum_{n\leq x}\frac{1}{n^2}+C/x=\sum_{n\leq x}\frac{1}{n^2}+\mathcal O(1/x). $$ But is there a constant independent of $x$ that makes this true?