teorema:
Si ( a_{n} >] 0 & m€N & ∑ ( m·a_{n} ) < oo ) ==> ∑ ( a_{n} ) < oo
demostració:
m·∑ ( a_{n} ) = ∑ ( m·a_{n} ) < oo
∑ ( a_{n} ) < (oo/m) < oo
teorema:
Si ( a_{n} >] 0 & ∑ ( oo·a_{n} ) < oo ) ==> ∑ ( a_{n} ) < 1
demostració:
oo·∑ ( a_{n} ) = ∑ ( oo·a_{n} ) < oo
∑ ( a_{n} ) < (oo/oo) = 1
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