advanced-calculus-evaluation-k-2-1-k-k-1-k-k-2-1-k-k-1-k-k-2-1-k-k-n-2-1-n-k-

Question Number 133068 by mnjuly1970 last updated on 18/Feb/21

\dots . a d v a n c e d \dots . . c a l c u l u s \dots .

e v a l u a t i o n :: \underset{k = 2}{\sum^{\infty}} {(- 1)}^{k} (\frac{ζ (k) - 1}{k})

::: Φ = \underset{k = 2}{\sum^{\infty}} {(- 1)}^{k} \frac{ζ (k) - 1}{k}

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

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

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

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

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

Φ_{n} = \underset{i = 1}{\sum^{n}} (\frac{1}{i + 1} - l n (1 + \frac{1}{i + 1}))

= (\underset{i = 1}{\sum^{n}} \frac{1}{i + 1}) - (l n (3) - l n (2) + l n (4) - l n (3) + . . + l n (n + 2) - l n (n + 1))

= l n (2) - l n (n + 2) + \underset{i = 1}{\sum^{n}} \frac{1}{i + 1}

= l n (2) - 1 + {\underset{i = 1}{\sum^{n}} \frac{1}{i} - l n (n + 2)}

∴ {l i m}_{n \to \infty} (Φ_{n}) = γ + 1 n (2) - 1

Φ = γ + l n (2) - 1 \dots . .

Commented by mnjuly1970 last updated on 18/Feb/21

m a m n o o n (g r a t e f u l) s i r p a y a n

Commented by Dwaipayan Shikari last updated on 18/Feb/21

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

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

= \underset{k = 2}{\sum^{\infty}} \frac{1}{k} - l o g (1 + \frac{1}{k})

= - 1 + l o g (2) + \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - l o g (\underset{k = 1}{\prod^{\infty}} \frac{k + 1}{k})

= - 1 + l o g (2) + \lim_{ϕ \to \infty} \underset{k = 1}{\sum^{\infty}} \frac{1}{k} - l o g (ϕ)

= γ - 1 + l o g (2)