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