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To-now-the-real-sequence-in-the-follwing-image-a-1-1-a-2-2-a-nk-1-a-2k-1-a-2k-2-k-Z-a-2k-2-a-2k-a-2k-1-then-prove-that-lim-n-a-n-3




Question Number 105130 by 175mohamed last updated on 26/Jul/20
To now the real sequence in the  follwing image :  a_1 =1   ,a_2 =2  a_(nk +1)  = ((a_(2k−1)  +a_(2k) )/2)           ∀ k ∈ Z^+   a_(2k+2)  = (√(a_(2k)  a_(2k+1) ))  then   prove that : lim_(n→∞)  a_n  = ((3(√3))/π)
$${To}\:{now}\:{the}\:{real}\:{sequence}\:{in}\:{the} \\ $$$${follwing}\:{image}\:: \\ $$$${a}_{\mathrm{1}} =\mathrm{1}\:\:\:,{a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{nk}\:+\mathrm{1}} \:=\:\frac{{a}_{\mathrm{2}{k}−\mathrm{1}} \:+{a}_{\mathrm{2}{k}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\forall\:{k}\:\in\:{Z}^{+} \\ $$$${a}_{\mathrm{2}{k}+\mathrm{2}} \:=\:\sqrt{{a}_{\mathrm{2}{k}} \:{a}_{\mathrm{2}{k}+\mathrm{1}} } \\ $$$${then}\: \\ $$$${prove}\:{that}\::\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\pi} \\ $$$$ \\ $$

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