teorema:
Si [∀k][ k€N ==> [∃s][ s > 0 & a_{k} [< s^{k} < 1 & (a_{k}/a_{k+1}) [< a_{k+(-1)} ] ] ==> ...
... ∑ a_{k} [< ( 1/( 1+(-1)·s^{2} ) ) [< oo
demostració:
(a_{k}/a_{k+1}) [< a_{k+(-1)}
a_{k} [< a_{k+1}a_{k+(-1)} [< s^{k+1}s^{k+(-1)} = s^{2k}
∑ a_{k} [< ∑ s^{2k} [< ( 1/( 1+(-1)·s^{2} ) ) [< oo
teorema:
Si [∀k][ k€N ==> [∃s][ s > 0 & a_{k} [< s^{k} < 1 & (a_{k}/a_{k+p}) [< a_{k+(-p)} ] ] ==> ...
... ∑ a_{k} [< ( 1/( 1+(-1)·s^{2} ) ) [< oo
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