S-2019-1-1-2-2-1-3-2-1-2019-2-

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

S_{2019} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{2019^{2}} = ?

Answered by Olaf_Thorendsen last updated on 21/Jun/21

S_{2019} = \underset{n = 1}{\sum^{2019}} \frac{1}{n^{2}}

S_{2019} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - \underset{n = 2020}{\sum^{\infty}} \frac{1}{n^{2}}

S_{2019} = ζ (2) - Ψ (1, 2020)

S_{2019} = \frac{π^{2}}{6} - Ψ (1, 2020)

S_{2019} \approx 1, 644438896

Commented by SOMEDAVONG last updated on 24/Jun/21

Thanks sir!!