Let's hope this attempt is better than the previous one.
[sp]$k^2>k^2-1 = (k-1)(k+1)$. Therefore $$\frac1{k^2} < \frac1{(k-1)(k+1)} = \frac1{2(k-1)} - \frac1{2(k+1)},$$ from which $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \sum_{k=n}^{2n-2}\Bigl(\frac1{2(k-1)} - \frac1{2(k+1)}\Bigr).$$ The sum on the right telescopes, leaving two orphan terms at the beginning and two at the end, to give $$\sum_{k=n}^{2n-2}\frac{1}{k^2}< \frac1{2(n-1)} + \frac1{2n} - \frac1{2(2n-2)} - \frac1{2(2n-1)} = \frac1{4(n-1)} + \frac1{2n} - \frac1{2(2n-1)}.$$ We want this to be less than (or equal to) $$\frac1{2n-1}$$, in other words $$\frac1{4(n-1)} + \frac1{2n} \leqslant \frac3{2(2n-1)}.$$ Multiply through by the denominators to see that this is equivalent to $$n(2n-1) + 2(n-1)(2n-1) \leqslant 6n(n-1).$$ That in turn is equivalent to $$\cancel{6n^2} -7n+2 \leqslant \cancel{6n^2} - 6n,$$ $$-n\leqslant-2.$$ Thus the given inequality holds provided that $n\geqslant2$.[/sp]