$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}∊(\frac{3}{2},\frac{7}{4}]$$ $$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}∊(\frac{3}{2},\frac{7}{4}]$$ $$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}∊(\frac{3}{2},\frac{7}{4}]$$