Σ_(k=1) ^(+∞) (−1)^k ln(1+(1/k))


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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
	$S=Σ_(k=1) ^n (−1)^k ∫_0 ^( 1) (dx/(x+k)) =∫_0 ^( 1) Σ_(k=1) ^∞ (−1)^k (1/(x+k))dx =_(convergent) ^(conditional) (1/2)∫_0 ^( 1) Σ(−1)^k (1/((x/2)+(k/2)))dx =(1/2)∫_0 ^( 1) {Σ_(k=) ^∞ (1/((x/2)+k)) −Σ_(k=1) ^∞ (1/((x/2)−(1/2)+k))}dx =(1/2)∫_0 ^( 1) {−γ+Σ(1/k)−(1/((x/2)−(1/2)+k))−(−γ+Σ(1/k)−(1/((x/2)+k)))}dx =(1/2)∫_0 ^( 1) {𝛙((x/2)+(1/2))−𝛙((x/2)+1)}dx note: 𝛙(z+1)=−𝛄+Σ_(k=1) ^∞ ((1/k)−(1/(k+z))) ∴ S=(1/2)(2ln(𝚪((x/2)+(1/2)))−2ln(𝚪((x/2)+1))_0 ^1 =−ln𝚪((3/2))−ln(𝚪((1/2)))= =−ln((1/2)(√π) .(√π) )=ln((2/π))...✓✓✓ =∫_0 ^( 1) ((1−x)/((1+x)ln(x)))dx .m.n.july.1970.$
	$S = \underset{k = 1}{\sum^{n}} {(- 1)}^{k} \int_{0}^{1} \frac{d x}{x + k}$ $= \int_{0}^{1} \underset{k = 1}{\sum^{\infty}} {(- 1)}^{k} \frac{1}{x + k} d x$ $\underset{c o n v e r g e n t}{\overset{c o n d i t i o n a l}{=}} \frac{1}{2} \int_{0}^{1} Σ {(- 1)}^{k} \frac{1}{\frac{x}{2} + \frac{k}{2}} d x$ $= \frac{1}{2} \int_{0}^{1} {\underset{k =}{\sum^{\infty}} \frac{1}{\frac{x}{2} + k} - \underset{k = 1}{\sum^{\infty}} \frac{1}{\frac{x}{2} - \frac{1}{2} + k}} d x$ $= \frac{1}{2} \int_{0}^{1} {- γ + Σ \frac{1}{k} - \frac{1}{\frac{x}{2} - \frac{1}{2} + k} - (- γ + Σ \frac{1}{k} - \frac{1}{\frac{x}{2} + k})} d x$ $= \frac{1}{2} \int_{0}^{1} {ψ (\frac{x}{2} + \frac{1}{2}) - ψ (\frac{x}{2} + 1)} d x$ $n o t e : ψ (z + 1) = - γ + \underset{k = 1}{\sum^{\infty}} (\frac{1}{k} - \frac{1}{k + z})$ $∴ S = \frac{1}{2} (2 l n (Γ (\frac{x}{2} + \frac{1}{2})) - 2 l n {(Γ (\frac{x}{2} + 1))}_{0}^{1}$ $= - l n Γ (\frac{3}{2}) - l n (Γ (\frac{1}{2})) =$ $= - l n (\frac{1}{2} \sqrt{π} . \sqrt{π}) = l n (\frac{2}{π}) . . . ✓ ✓ ✓$ $= \int_{0}^{1} \frac{1 - x}{(1 + x) l n (x)} d x . m . n . j u l y . 1970 .$
	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

	$g r a t e f u l s i r . . . . .$
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