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Calculus/testing for convergence/divergence

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Expert: - 3/29/2010


Question
I need to test the following for convergence/divergence for Sum n=3 to infinity:
(n+1)/(n^3*ln(n+2))
how do I go about doing this? Can I turn this into:
(1/n^2+1/n^3)/(ln(n+2)) and use L'Hopital's rule?
Or do I need to use an Integral test? What do i make u=?
Thanks!

Answer
I would divide the fraction into two pieces.
That is, make it into n/(n^3*ln(n+2)) + 1/(n^3*ln(n+2)).

In the first fraction, n/n cancels, leaving 1/(n²*ln(n+2)) as the 1st fraction.

Note that in the 1st fraction, each term is less that 1/n², and that converges.
Note that in the 2nd fraction, each term is less than 1/n³, and that converges.

Since both terms converge, the entire expression converges.

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