teorema:
Si ( a_{n} >] 0 & b_{n} >] 0 & ∑ ( a_{n} ) < oo & ∑ ( b_{n} ) < oo ) ==> ...
... ∑ ( (a_{n}+b_{n})/2 ) < oo
demostració:
∑ ( a_{n}+b_{n} ) = ∑ ( a_{n} ) + ∑ ( b_{n} ) < 2·oo
teorema:
Si ( a_{n} >] 0 & b_{n} >] 0 & ∑ ( a_{n} ) < oo & ∑ ( b_{n} ) < oo ) ==> ...
... ∑ ( (a_{n}·b_{n})^{(1/2)} ) < oo
demostració:
∑ ( (a_{n}·b_{n})^{(1/2)} ) [< ∑ ( (a_{n}+b_{n})/2 ) < oo
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