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1-1-2-3-2-1-2-3-4-2-1-3-4-5-2-1-16-4pi-2-39-




Question Number 141577 by qaz last updated on 20/May/21
((1/(1∙2∙3)))^2 +((1/(2∙3∙4)))^2 +((1/(3∙4∙5)))^2 +...=(1/(16))(4π^2 −39)
$$\left(\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}\right)^{\mathrm{2}} +…=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{4}\pi^{\mathrm{2}} −\mathrm{39}\right) \\ $$

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