→ yamadahosi :謝謝 10/25 19:19
Because a_nb_n is converge, a_nb_n< M for some M>0.
It implies b_n<M/a_n
Now claim: Given e>0 , there exists N such that b_n<M/a_n<e for all n>N.
Clearly, we can find suitable M' such that M/M'<e.
Since a_n is diverge to infinite, then there exists
an N such that |a_n|>M' for all n>N.
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