n-2-3-3n-1-

Question Number 116057 by Study last updated on 30/Sep/20

\underset{n = 2}{\sum^{\infty}} \frac{3}{3 n + 1} = ?

Answered by mindispower last updated on 30/Sep/20

\sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(3 n + 1)} ?

Commented by mnjuly1970 last updated on 01/Oct/20

a n s w e r ::= \frac{π}{3 \sqrt{3}} + \frac{l n (2)}{3} ✓

Commented by mnjuly1970 last updated on 01/Oct/20

\frac{1}{2} \underset{n = - \infty}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(3 n + 1)} \neq \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{3 n + 1}

l . h . s = \frac{1}{2} [\dots . . + \frac{1}{14} - \frac{1}{11} + \frac{1}{8} - \frac{1}{5} + \frac{1}{2} + \frac{1}{1} - \frac{1}{4} + + \frac{1}{7} - \frac{1}{10} + \dots

Commented by mnjuly1970 last updated on 01/Oct/20

b e c a u s e

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{3 n + 1} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{3 n} d x

= \int_{0}^{1} \frac{1}{1 + x^{3}} d x = \frac{π}{3 \sqrt{3}} + \frac{l n (2)}{3} ✓

. m . n . j u l y . 70

Answered by mathmax by abdo last updated on 30/Sep/20

this serie is divergent due to \frac{3}{3 n + 1} \sim \frac{1}{n}