Since (bn) is bounded, there exist k >0 such that mod bn <=k for all n.
Therefore, mod anbn <= k mod an
Now, let excellon >0 be given.
Since, (an) ->0, there exists m belongs to N such that mod an < excellon/k for all n >=m.
Therefore, mod anbn < excellon for all n >=m.
Therefore, (anbn) ->0.
Therefore, mod anbn <= k mod an
Now, let excellon >0 be given.
Since, (an) ->0, there exists m belongs to N such that mod an < excellon/k for all n >=m.
Therefore, mod anbn < excellon for all n >=m.
Therefore, (anbn) ->0.
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