sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn]
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Originally Posted by twilightstr sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn] Do you mean ? An obvious first step would be to multiply numerator and denominator by so that each term is . Now a comparison test should do it. Or, now that I think about it, just a comparison to !
no. 1/[(squareroot (n+1)) - (squareroot (n-1))]
Originally Posted by twilightstr no. 1/[(squareroot (n+1)) - (squareroot (n-1))] If you mean then HallsofIvy's suggestion still works - multiply btop and bottom by giving which diverges (terms don't go to zero) You could also do a limit comparison with
Since the second is divergent (You should be able to figure that is divergent). By the comparison test, must also diverge.
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