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