400 Followers Problem - Reciprocal Summations 2

Wait it's not tedious :

S can be written as :

$\displaystyle S = \sum _{ r=0 }^{ \infty }{ \frac { { (-1) }^{ r } }{ 3r+1 } } +\sum _{ r=0 }^{ \infty }{ \frac { { (-1) }^{ r } }{ 3r+2 } } +\sum _{ r=0 }^{ \infty }{ \frac { { (-1) }^{ r } }{ 3r+2 } }$

$\Rightarrow \displaystyle S = \sum _{ r=0 }^{ \infty }{ { (-1) }^{ r }\int _{ 0 }^{ 1 }{ { x }^{ 3r }dx } } +\sum _{ r=0 }^{ \infty }{ { (-1) }^{ r }\int _{ 0 }^{ 1 }{ { x }^{ 3r+1 }dx } } +\sum _{ r=0 }^{ \infty }{ { (-1) }^{ r }\int _{ 0 }^{ 1 }{ { x }^{ 3r+2 }dx } }$

Interchanging summation and integral we have :

$\displaystyle S = \int _{ 0 }^{ 1 }{ (1+x+{ x }^{ 2 })\sum _{ r=0 }^{ \infty }{ { (-1) }^{ r }({ x }^{ 3r }) } dx }$

$\Rightarrow \displaystyle S = \int _{ 0 }^{ 1 }{ \frac { (1+x+{ x }^{ 2 }) }{ 1+{ x }^{ 3 } } dx }$

Doing this is easy :

$\displaystyle S = \int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } +\int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }-x+1 } }$

These integrals are quite trivial and hence we have the answer as :

$\displaystyle S = \left( \frac { 1 }{ 3 } \ln { ({ x }^{ 3 }+1) } +\frac { 2 }{ \sqrt { 3 } } \arctan { \left( \frac { 2x-1 }{ \sqrt { 3 } } \right) } \right) { | }_{ 0 }^{ 1 }=\frac { 2\sqrt { 3 } \pi }{ 9 } +\frac { \ln { (2) } }{ 3 }$

By observing that after every 6 terms, the sign comes back, we can see

$\displaystyle S = \sum_{k=0}^{\infty}{\frac{1}{6k+1}} + \sum_{k=0}^{\infty}{\frac{1}{6k+2}} + \sum_{k=0}^{\infty}{\frac{1}{6k+3}} - \sum_{k=0}^{\infty}{\frac{1}{6k+4}} - \sum_{k=0}^{\infty}{\frac{1}{6k+5}} - \sum_{k=0}^{\infty}{\frac{1}{6k+6}}$

A really simple way would be to use digamma function and I will do the easiest.

We know $\displaystyle \psi(s+1) = \int_{0}^{1}{\frac{1-x^s}{1-x} \ dx} - \gamma$

$\displaystyle \sum_{k=0}^{\infty}{\frac{1}{6k+m+1}} = \int_{0}^{1}{\frac{x^m}{1-x^6} \ dx}$

$\displaystyle \sum_{k=0}^{\infty}{\frac{1}{6k+1}} + \sum_{k=0}^{\infty}{\frac{1}{6k+2}} + \sum_{k=0}^{\infty}{\frac{1}{6k+3}} - \sum_{k=0}^{\infty}{\frac{1}{6k+4}} - \sum_{k=0}^{\infty}{\frac{1}{6k+5}} - \sum_{k=0}^{\infty}{\frac{1}{6k+6}} =$

$\displaystyle \frac{1}{6} \left(\int_{0}^{1}{\frac{{t}^{-5/6}}{1-t} + \frac{{t}^{-4/6}}{1-t} + \frac{{t}^{-3/6}}{1-t} - \frac{{t}^{-2/6}}{1-t} - \frac{{t}^{-1/6}}{1-t} - \frac{{t}^{-0/6}}{1-t} \ dt}\right)$

$\displaystyle = -\frac{1}{6}\left(\psi\left(1 -\frac{5}{6}\right) + \psi\left(1 -\frac{4}{6}\right) + \psi\left(1 -\frac{3}{6}\right) - \psi\left(1 -\frac{2}{6}\right) - \psi\left(1 -\frac{1}{6}\right) - \psi\left(1 -\frac{0}{6}\right)\right)$

$\displaystyle = \frac{1}{6}\left(\psi\left(1\right) + \psi\left(\frac{5}{6}\right) + \psi\left(\frac{4}{6}\right) - \psi\left(\frac{3}{6}\right) - \psi\left(\frac{2}{6}\right) - \psi\left(\frac{1}{6}\right)\right)$

$\displaystyle = \frac{2\sqrt{3}}{9}\pi + \frac{\ln(2)}{3}$

The sum is equal to S= x +y + z, where x = 1 - 1/4 + 1/7- 1/10 + 1/13 - ...=pi.sqrt(3)/9 + ln(2)/3 , y = 1/2 - 1/5 + 1/8 - 1/11 + ... = pi.sqrt(3)/9 - ln(2)/3 , z = 1/3 - 1/6 +1/9 - 1/12 + ... = ln(2)/3. So S = 2.pi.sqrt(3)/9 + ln(2)/3.