# Video: Applying the πth Term Divergence Test with a Trigonometric Function

What can we conclude by applying the πth term divergence test in the series β _(π = 1) ^(β) 3 cos π?

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### Video Transcript

What can we conclude by applying the πth term divergence test in the series the sum from π equals one to β of three cos π?

Letβs begin by recalling what the πth term divergence test tells us. It tells us that if the limit as π approaches β of π sub π does not exist or if itβs not equal to zero, then the series the sum from π equals one to β of π sub π is divergent. We also recall that if the limit is equal to zero, we canβt tell whether the series converges or diverges. And so we say that the test fails or itβs inconclusive.

In this question, we can define π sub π to be equal to three cos π. And weβre going to need to evaluate the limit as π approaches β of three cos π. Now actually, thereβs not a lot we need to do to evaluate this limit. Letβs sketch the graph of three cos π and see if we can recall what it looks like.

The graph of π¦ equals three cos π₯ is shown. We can see that itβs an oscillating function. It has peaks and troughs at π¦-values of three and negative three. This continues forever. And so we say that the limit as π approaches β of three cos π does not exist. We cannot pinpoint it as one particular number or even β.

And by the πth term divergence test, that tells us that the series the sum from π equals one to β of three cos π is divergent. It diverges.