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2/(n2+3)^1/2 , Any Idea?

The nth term of the sequence is given; check whether the sequence is convergent or divergent.

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    Each term of the series whose nth term is 2/sqrt[n^2+3] is equal to or larger than each term of the series 1/n, the sum of which does not converge. Therefore, the series in question does not have a sum that converges.
    1 0

    Oddman 

    answered 1 year ago

      2/(n2+3)^1/2

      If we put n= infinity then the net result is 0, so the
      sequence converges to zero.
      0 0

      Anukhan 

      answered 1 year ago

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