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nikitaaz
@nikitaaz
June 2022
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По признаку Даламбера
lim(n->oo) a(n+1) / a(n) =
= lim(n->oo) [4^(n+1)*(n+2)! / (2n+2)!] / [4^n*(n+1)! / (2n)!] =
= lim(n->oo) [4*4^n*(n+2)(n+1)! / ((2n+2)(2n+1)(2n)!)] * [(2n)! / (4^n*(n+1)!)] =
= lim(n->oo) [4*(n+2)] / [(2n+2)(2n+1)] = lim(n->oo) (4n+8) / (4n^2+6n+2) = 0
Этот предел меньше 1, значит, ряд сходится.
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Answers & Comments
Verified answer
По признаку Даламбераlim(n->oo) a(n+1) / a(n) =
= lim(n->oo) [4^(n+1)*(n+2)! / (2n+2)!] / [4^n*(n+1)! / (2n)!] =
= lim(n->oo) [4*4^n*(n+2)(n+1)! / ((2n+2)(2n+1)(2n)!)] * [(2n)! / (4^n*(n+1)!)] =
= lim(n->oo) [4*(n+2)] / [(2n+2)(2n+1)] = lim(n->oo) (4n+8) / (4n^2+6n+2) = 0
Этот предел меньше 1, значит, ряд сходится.