calculation Ω= Σ_(n=1) ^∞ (( ζ ( 2n ))/4^( n) ) =^? (1/2) Ω =Σ_(n=1) ^∞ {(1/2^( 2n) ) Σ_(k=1) ^∞ (1/k^( 2n) ) } = Σ_(n=1) ^∞ Σ_(k=1) ^∞ (1/((2k )^( 2n) ))=Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((2k)^( 2n) )) = Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((4k^( 2) )^( n) )) = Σ_(k=1) ^∞ (( (1/(4k^( 2) )))/(1− (1/(4k^( 2) )))) = Σ_(k=1) ^∞ (1/((2k−1)(2k+1 ))) =(1/2) Σ_(k=1) ^∞ ((1/(2k−1)) −(1/(2k+1)) ) = (1/2) (1/(2(1)−1)) − lim


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Question Number 180236 by mnjuly1970 last updated on 09/Nov/22
$calculation Ω= Σ_(n=1) ^∞ (( ζ ( 2n ))/4^( n) ) =^? (1/2) Ω =Σ_(n=1) ^∞ {(1/2^( 2n) ) Σ_(k=1) ^∞ (1/k^( 2n) ) } = Σ_(n=1) ^∞ Σ_(k=1) ^∞ (1/((2k )^( 2n) ))=Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((2k)^( 2n) )) = Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((4k^( 2) )^( n) )) = Σ_(k=1) ^∞ (( (1/(4k^( 2) )))/(1− (1/(4k^( 2) )))) = Σ_(k=1) ^∞ (1/((2k−1)(2k+1 ))) =(1/2) Σ_(k=1) ^∞ ((1/(2k−1)) −(1/(2k+1)) ) = (1/2) (1/(2(1)−1)) − lim_( k→∞) (1/(2k+1)) =(1/2)$
$calculation$ $Ω = \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n)}{4^{n}} \overset{?}{=} \frac{1}{2}$ $Ω = \underset{n = 1}{\sum^{\infty}} {\frac{1}{2^{2 n}} \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2 n}}}$ $= \underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{1}{{(2 k)}^{2 n}} = \underset{k = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 k)}^{2 n}}$ $= \underset{k = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(4 k^{2})}^{n}} = \underset{k = 1}{\sum^{\infty}} \frac{\frac{1}{4 k^{2}}}{1 - \frac{1}{4 k^{2}}}$ $= \underset{k = 1}{\sum^{\infty}} \frac{1}{(2 k - 1) (2 k + 1)} = \frac{1}{2} \underset{k = 1}{\sum^{\infty}} (\frac{1}{2 k - 1} - \frac{1}{2 k + 1})$ $= \frac{1}{2} \frac{1}{2 (1) - 1} - \lim_{k \to \infty} \frac{1}{2 k + 1} = \frac{1}{2}$
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