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Question Number 38195 by maxmathsup by imad last updated on 22/Jun/18

l e t x ⩾ 1 a n d δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}}

1) c a l c u l a t e δ (x) i n t e r m s o f ξ (x) i f x > 1

2) f i n d δ (1)

3) f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

4) c a l c u l a t e δ (3) i n t e r m s o f ξ (3) .

Commented by math khazana by abdo last updated on 23/Jun/18

1) δ (x) = \sum_{n = 1}^{\infty} \frac{1}{{(2 n)}^{x}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}}

= 2^{- x} ξ (x) - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}} b u t w e h a v e

ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}} = \sum_{n = 1}^{\infty} \frac{1}{{(2 n)}^{x}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}}

= 2^{- x} ξ (x) + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{x}} = (1 - 2^{- x}) ξ (x) \Rightarrow

δ (x) = 2^{- x} ξ (x) - (1 - 2^{- x}) ξ (x)

= (2^{1 - x} - 1) ξ (x)

2) δ (1) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)

3) w e h a v e \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = (1 - 2^{- 1}) ξ (2)

= \frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8} .

4) δ (3) = (2^{1 - 3} - 1) ξ (3) = (\frac{1}{4} - 1) ξ (3)

= - \frac{3}{4} ξ (3) \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3}} = - \frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{3}} .