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Question Number 74863 by mrS last updated on 02/Dec/19 Commented by abdomathmax last updated on 03/Dec/19 $${we}\:{have}\:{S}=\sum_{{n}=\mathrm{1}} ^{\mathrm{45}} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{\mathrm{45}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{\mathrm{45}−\mathrm{1}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}}…
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