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Question Number 144068 by SOMEDAVONG last updated on 21/Jun/21

S_(2019) =1+ (1/2^2 ) + (1/3^2 ) + ...+ (1/(2019^2 )) =?

$$\mathrm{S}_{\mathrm{2019}} =\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:...+\:\frac{\mathrm{1}}{\mathrm{2019}^{\mathrm{2}} }\:=? \\ $$

Answered by Olaf_Thorendsen last updated on 21/Jun/21

S_(2019)  = Σ_(n=1) ^(2019) (1/n^2 )  S_(2019)  = Σ_(n=1) ^∞ (1/n^2 )−Σ_(n=2020) ^∞ (1/n^2 )  S_(2019)  = ζ(2)−Ψ(1,2020)  S_(2019)  = (π^2 /6)−Ψ(1,2020)  S_(2019)  ≈ 1,644438896

$$\mathrm{S}_{\mathrm{2019}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2019}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\mathrm{S}_{\mathrm{2019}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\underset{{n}=\mathrm{2020}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\mathrm{S}_{\mathrm{2019}} \:=\:\zeta\left(\mathrm{2}\right)−\Psi\left(\mathrm{1},\mathrm{2020}\right) \\ $$$$\mathrm{S}_{\mathrm{2019}} \:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\Psi\left(\mathrm{1},\mathrm{2020}\right) \\ $$$$\mathrm{S}_{\mathrm{2019}} \:\approx\:\mathrm{1},\mathrm{644438896} \\ $$

Commented by SOMEDAVONG last updated on 24/Jun/21

Thanks sir!!

$$\mathrm{Thanks}\:\mathrm{sir}!! \\ $$

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