k-1-1-k-ln-1-1-k-

Question Number 132928 by metamorfose last updated on 17/Feb/21

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

Commented by Olaf last updated on 17/Feb/21

s o r r y s i r, I d e l e t e d m y a n s w e r .

i t w a s w r o n g .

Answered by mnjuly1970 last updated on 17/Feb/21

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

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

\underset{c o n v e r g e n t}{\overset{c o n d i t i o n a l}{=}} \frac{1}{2} \int_{0}^{1} Σ {(- 1)}^{k} \frac{1}{\frac{x}{2} + \frac{k}{2}} d x

= \frac{1}{2} \int_{0}^{1} {\underset{k =}{\sum^{\infty}} \frac{1}{\frac{x}{2} + k} - \underset{k = 1}{\sum^{\infty}} \frac{1}{\frac{x}{2} - \frac{1}{2} + k}} d x

= \frac{1}{2} \int_{0}^{1} {- γ + Σ \frac{1}{k} - \frac{1}{\frac{x}{2} - \frac{1}{2} + k} - (- γ + Σ \frac{1}{k} - \frac{1}{\frac{x}{2} + k})} d x

= \frac{1}{2} \int_{0}^{1} {ψ (\frac{x}{2} + \frac{1}{2}) - ψ (\frac{x}{2} + 1)} d x

n o t e : ψ (z + 1) = - γ + \underset{k = 1}{\sum^{\infty}} (\frac{1}{k} - \frac{1}{k + z})

∴ S = \frac{1}{2} (2 l n (Γ (\frac{x}{2} + \frac{1}{2})) - 2 l n {(Γ (\frac{x}{2} + 1))}_{0}^{1}

= - l n Γ (\frac{3}{2}) - l n (Γ (\frac{1}{2})) =

= - l n (\frac{1}{2} \sqrt{π} . \sqrt{π}) = l n (\frac{2}{π}) \dots ✓ ✓ ✓

= \int_{0}^{1} \frac{1 - x}{(1 + x) l n (x)} d x . m . n . j u l y . 1970 .

Commented by metamorfose last updated on 17/Feb/21

t h a n k s, {i t}^{'} s a n i n t r e s t i n g s o l u t i o n

Commented by mnjuly1970 last updated on 17/Feb/21

g r a t e f u l s i r \dots . .