Does the function converge or diverge #1/((x-1)(x^2+1)) #on the bound [2,infinity]?

1 Answer
MoominDave
Jun 2, 2018

Assuming poles are meant, #x=+-1#

Explanation:

The nomenclature is a bit unclear - the given function doesn't really "converge" or "diverge" in the way it is specified - these are more properties of sequences or series. Functions can be specified as series, a manner in which they can converge or diverge for various values, but that doesn't seem to apply here.

I'm going to assume that what is meant is to find the #x# values for which the function blows up to infinity. These occur when the denominator is equal to zero. In this case the denominator has handily already been factored for us, and has three roots with two distinct values, #+-1#.

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