calculate-k-2-1-k-k-k-if-s-n-1-1-n-s-

Question Number 66483 by ~ À ® @ 237 ~ last updated on 16/Aug/19

c a l c u l a t e \underset{k = 2}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k} ζ (k) i f ζ (s) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{s}}

Commented by mathmax by abdo last updated on 16/Aug/19

l e t S = \sum_{k = 2}^{\infty} \frac{{(- 1)}^{k}}{k} ξ (k) \Rightarrow S = \sum_{k = 2}^{\infty} \frac{{(- 1)}^{k}}{k} (\sum_{n = 1}^{\infty} \frac{1}{n^{k}})

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

l e t f i n d f (x) = \sum_{k = 2}^{\infty} \frac{x^{k}}{k} w i t h ∣ x ∣< 1 \Rightarrow f^{^{'}} (x) = \sum_{k = 2}^{\infty} x^{k - 1} = \sum_{k = 1}^{\infty} x^{k}

= x \sum_{k = 1}^{\infty} x^{k - 1} [= x \sum_{k = 0}^{\infty} x^{k} = \frac{x}{1 - x} \Rightarrow f (x) = \int \frac{x}{1 - x} d x + c

= - \int \frac{x}{x - 1} d x + c = - \int \frac{x - 1 + 1}{x - 1} d x + c = - x + l n ∣ x - 1 ∣ + c

f (0) = 0 = c \Rightarrow f (x) = - x + l n ∣ x - 1 ∣\Rightarrow

\sum_{k = 2}^{\infty} \frac{1}{k} {(- \frac{1}{n})}^{k} = \frac{1}{n} + l n ∣ \frac{1}{n} + 1 ∣ = \frac{1}{n} + l n (\frac{n + 1}{n})

= \frac{1}{n} + l n (n) - l n (n + 1) \Rightarrow

S = \sum_{n = 1}^{\infty} (\frac{1}{n} + l n (n) - l n (n + 1)) = {l i m}_{n \to + \infty} S_{n} w i t h

S_{n} = \sum_{k = 1}^{n} (\frac{1}{k} + l n (k) - l n (k + 1)) = \sum_{k = 1}^{n} \frac{1}{k} + \sum_{k = 1}^{n} (l n (k) - l n (k + 1))

= H_{n} - l n (n + 1) = H_{n} - l n (n) - l n (1 + \frac{1}{n}) \to γ (n \to + \infty) \Rightarrow

\sum_{k = 2}^{\infty} \frac{{(- 1)}^{k}}{k} ξ (k) = γ (c o n s t a n t o f E u l e r)

Commented by mathmax by abdo last updated on 17/Aug/19

f (x) = - x - l n ∣ x - 1 ∣\Rightarrow \sum_{k = 2}^{\infty} \frac{1}{k} {(- \frac{1}{n})}^{k} = \frac{1}{n} - l n (\frac{1}{n} + 1)

= \frac{1}{n} - l n (\frac{n + 1}{n}) = \frac{1}{n} + l n (n) - l n (n + 1) \dots (e r r o r o f t y p o)

Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 17/Aug/19

y o u a r e w e l c o m e .