find-the-partial-sums-of-n-1-1-n-2-n-1-

Question Number 143131 by mohammad17 last updated on 10/Jun/21

f i n d t h e p a r t i a l s u m s o f \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1)}

Answered by qaz last updated on 10/Jun/21

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1)}

= \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{2}} dx

= \int_{0}^{1} {Li}_{2} (x) dx

= {xLi}_{2} (x) ∣_{0}^{1} + \int_{0}^{1} \frac{xln (1 - x)}{x} dx

= \frac{π^{2}}{6} + \int_{0}^{1} \ln (1 - x) dx

= \frac{π^{2}}{6} - 1

Commented by mohammad17 last updated on 11/Jun/21

t h a n k y o u s i r b u t w h y t h e i n t e r v a l o f t h e

i n t e g r a l f r o m (0 t o 1)

Answered by Olaf_Thorendsen last updated on 10/Jun/21

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1)}

S = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n^{2}} - \frac{1}{n} + \frac{1}{n + 1})

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1})

S = \frac{π^{2}}{6} - 1