and both have a pole at
From the Laurent series expansion I deduced that for this pole is of order one, while for the pole is order two.
Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
... where is . An interesting 'counterexample' is the function , for which is a 'singularity'. If we try to apply (1) in order to determine the 'order of the pole' we discover that...
... so that no value on n satisfies (1). The reason is that is a singularity of the function but isn't a 'pole'...