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In this video, weβre going to learn about something called the πth term divergence test.
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Weβll learn how to use it to prove that a series diverges and look at what happens when the test fails.
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Weβll also consider what this test can tell us about a convergent series.
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We begin by recalling that a sequence is called convergent if the limit of the sum of its first π terms as π approaches infinity is equal to some finite number π , known as the sum of the series.
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Similarly, the sequence is divergent if the converse is true, if the sum of the series does not have a finite limit.
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So how do we test for this?
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Well, there are a number of methods.
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In this video, weβre going to use something called the πth term divergence test to establish whether a series diverges or whether the test fails.
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The first definition weβre interested in is that if the series the sum of ππ from π equals one to infinity is convergent, then the limit as π approaches infinity of ππ is equal to zero.
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And itβs really important to realise that the converse of this theorem is not necessarily true.
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If the limit as π approaches infinity of ππ is equal to zero, we canβt conclude that the series is convergent.
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For example, the harmonic series the sum of one over π, we have ππ equals one over π, which approaches zero as π approaches infinity.
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But that is a divergent series.
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And so we come to the πth term divergence test.
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This says if the limit as π approaches infinity does not exist or is not equal to zero, then the series the sum of ππ from π equals one to infinity is divergent.
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Notice that, once again, this shows us that if the limit is equal to zero, we know nothing about the convergence or divergence of this series.
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Itβs really important that we realise that when the πth term divergence test gives an answer of zero, we say that the test fails.
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Letβs now consider an example.
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Using the πth term test, determine whether the series the sum of π over π squared plus one from π equals zero to infinity is divergent or the test fails.
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Remember, the πth term test for divergence says that if the limit as π approaches infinity of ππ does not exist or if the limit is not equal to zero, then the series the sum of ππ from π equals one to infinity is divergent.
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We also recall that if the 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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Now notice the sum in our question is from π equals zero to infinity, as opposed to from π equals one to infinity.
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In practice, this doesnβt really matter.
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Weβre looking for absolute divergence, essentially whatβs happening to the series as π gets larger and larger.
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And if we look carefully, we see that when π equals zero, our first term is also zero.
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So we can split this into zero plus the sum from π equals one.
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In our case then, weβre going to let ππ be equal to π over π squared plus one.
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And weβre going to evaluate the limit as π approaches infinity of π over π squared plus one.
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And when evaluating a limit, we should always check whether we can use direct substitution.
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In this case, if we were to substitute π equals infinity into the expression, we get infinity over infinity, which we know to be indeterminate.
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So instead, we look for a way to manipulate the expression π over π squared plus one.
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We do so by dividing both the numerator and denominator by π squared.
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Remember, we can do this as it creates an equivalent fraction.
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And we choose π squared as itβs the highest power of π in our denominator.
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So we get the limit as π approaches infinity of π over π squared over π squared over π squared plus one over π squared.
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This simplifies to the limit as π approaches infinity of one over π over one plus one over π squared.
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Weβre then going to use the division law for limits.
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This says that the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π of π₯ over the limit as π₯ approaches π of π of π₯.
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And thatβs provided the limits exist and the limit of π of π₯ is not equal to zero.
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And so this becomes the limit as π approaches infinity of one over π over the limit as π approaches infinity of one plus one over π squared.
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And we can now substitute π equals infinity into this expression.
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As π gets larger, one over π gets smaller.
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Ultimately, it approaches zero.
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Similarly, as π gets larger, one over π squared also approaches zero.
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And so our limit becomes zero over one plus zero, which is of course simply zero.
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And so since the limit as π approaches infinity of ππ is equal to zero, we see that the test fails.
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Weβll now consider another example.
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What can we conclude by applying the πth term divergence test in the series the sum of three π over the square root of six π squared plus four π plus five for values of π from one to infinity?
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Remember, the πth term test for divergence says that if the limit as π approaches infinity of ππ does not exist or the limit is not equal to zero, the series the sum of ππ from π equals one to infinity is divergent.
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And we recall that if the 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 this question, weβre going to let ππ be equal to three π over the square root of six π squared plus four π plus five.
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And so we need to evaluate the limit of this expression as π approaches infinity.
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We canβt use direct substitution.
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If we were to do so, we would get infinity over infinity, which we know is indeterminate.
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So instead, we need to find a way to manipulate our expression and see if that will help us to evaluate the limit.
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To make our life just a little bit easier, letβs apply the constant multiple rule.
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And this is that the limit as π approaches infinity of some constant times the function in π is equal to that constant times the limit of the function in π.
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So essentially, we can take the constant three outside of our limit.
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And then this next step is going to appear a little bit strange.
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Weβre going to divide both the numerator and denominator of our expression by π.
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The numerator then becomes one.
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And the denominator becomes one over π times the square root of six π squared plus four π plus five.
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Weβre now going to take this one over π inside our square root.
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To do that, weβll need to square it.
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So the denominator becomes the square root of six π squared over π squared plus four π over π squared plus five over π squared.
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And then, actually, this simplifies quite nicely.
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We now have three times the limit as π approaches infinity of one over the square root of six plus four over π plus five over π squared.
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And weβre now ready to apply direct substitution.
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As π gets larger, four over π and five over π squared get smaller.
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They approach zero.
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One and six are both independent of π.
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So our limit becomes one over the square root of six.
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We want to rationalise the denominator here.
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So we multiply both the numerator and denominator of our fraction by the square root of six.
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And we see that the limit as π approaches infinity of ππ is three times the square root of six over six, which is root six over two.
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We notice that this is not equal to zero.
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And this leads us to conclude that the series the sum of three π over the square root of six π squared plus four π plus five between π equals one and infinity is divergent.
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Weβll now consider a slightly more complicated example that requires use of an additional rule for finding limits.
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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.
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In this video, weβve seen that the πth term divergence test can tell us whether a series is divergent.
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And it tells us 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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We saw that if the series the sum of ππ from π equals one to infinity is convergent, then the limit as π approaches infinity of ππ will be zero.
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But that itβs really important to realise that the converse of this theorem is not necessarily true.
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If the limit as π approaches infinity of π of π is equal to zero, we canβt conclude that the series is convergent.
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And in fact, we say that the test fails.