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What can we conclude by applying the πth term divergence test in the series the sum of two times the natural log of π over three π for π equals one to infinity?
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We begin by recalling that the πth term test for divergence says that if the limit as π approaches infinity of ππ is not equal to zero or does not exist, then the series the sum of ππ from π equals one to infinity is divergent.
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And indeed, if that limit is equal to zero, we canβt tell whether the series converges or diverges and we say that the test fails.
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In our question then, weβre going to let ππ be equal to two times the natural log of π over three π.
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And so our job is to evaluate the limit as π approaches infinity of this expression.
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If we were to simply apply direct substitution, then weβd find that our limit is equal to infinity over infinity.
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And of course, thatβs indeterminate.
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So instead, weβre going to recall LβHΓ΄pitalβs rule.
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This says if the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to infinity over infinity, then the limit as π₯ approaches π of π prime of π₯ over π prime of π₯ will tell us the value of the limit as π₯ approaches π of π of π₯ over π of π₯.
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We can also use this formula if our limit is equal to zero over zero.
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But weβre not interested in that case.
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Now of course, weβre working with π.
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So weβre going to need to differentiate two times the natural log of π and three π with respect to π.
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The derivative of the natural log of π is one over π.
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So when we differentiate two times the natural log of π with respect π, we get two over π.
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And then the derivative of three π is simply three.
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So we can now evaluate this as π approaches infinity.
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As π gets larger, two over π gets smaller.
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And as π approaches infinity therefore, two over π approaches zero.
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We find that this is therefore equal to zero over three, which is zero.
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And we find that the test fails or itβs inconclusive.