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No. The sequence (n*sin(n)) is not properly divergent. To be properly divergent it must either "tend to" +inf or -inf. We say that (xn) tends to +inf if for every real number a there exists a natural number N such that if n>=N, then xn>a.
It is clear that no such N exists for all real numbers because n*sin(n) oscillates (because of the sin(n)). Therefore (n*sin(n)) is not properly divergent. This is not a rigorous proof but the definition of proper divergence is precise and can be used for any proof dealing with proper divergence.
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It is called a divergent plate boundary.
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Not always true. Eg the divergent series 1,0,2,0,3,0,4,... has both convergent and divergent sub-sequences.
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Divergent has been challenged.
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