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# Determine the convergence or divergence of f(n) = ne-πin / 2

Consider the function defined by

Determine whether converges or diverges, and if it converges find its limit.

This sequence diverges.
Proof. We saw in this exercise (Section 10.4, Exercise #20) that the sequence defined by diverges. We could use this to show that this sequence diverges (since for we have and so cannot approach a finite limit). For practice, we can also prove this directly from the definition by contradiction as follows. Suppose there exists a real number and a positive integer such that for all and all we have

Since is positive we know and . So, letting , we have

Adding these two expressions and using the triangle inequality we have,

This is a contradiction. Hence, the sequence does not tend to a limit .

### One comment

1. William C says:

It mightv’e been simpler to look at the magnitudes of n and e^(-piin/2). Since the first is constant at one and the other approaches infinity, the overall magnitude approaches infinity, implying the limit approaches infinity and the sequence diverges (someone let me know if this doesn’t work for some reason)