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Consider the power series the sum of three π₯ to the πth power over π plus five for π equals zero to β.
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Find the interval of convergence of the power series and find the radius of convergence of the power series.
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To test for convergence of our power series, weβre going to recall the ratio test.
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The part of the ratio test weβre interested in says that, given a series the sum of π π, if the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent.
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If we look at our series, we see that we can define π π as three π₯ to the πth power over π plus five.
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And then π π plus one is three π₯ to the power of π plus one over π plus one plus five.
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And then of course that denominator simplifies nicely to π plus six.
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Our job is to establish where the limit as π approaches β of the absolute value of the quotient of these is less than one.
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And we know that when dividing by a fraction, we multiply by the reciprocal of that very same fraction.
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So the inside of our limit becomes three π₯ to the power of π plus one over π plus six times π plus five over three π₯ to the πth power.
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And we now see that we can divide through by three π₯ to the πth power.
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Okay, so now we have the limit as π approaches β of the absolute value of three π₯ times π plus five over π plus six.
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We notice that three π₯ is independent of π.
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So we can take the absolute value of three π₯ outside our limit.
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And then we divide through both the numerator and denominator of the fraction by π.
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Thatβs the highest power of π in our denominator.
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And so the denominator becomes π divided by π, which is one, plus five over π.
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And then the numerator becomes one plus six over π.
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And as π approaches β, five over π and six over π approach zero.
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And so the limit as π approaches β of the absolute value of one plus five over π over one plus six over π is simply the absolute value of one or one.
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So we need to establish where the absolute value of three π₯ times one is less than one, or simply where the absolute value of three π₯ is less than one.
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Since three is a purely positive number, we can divide through both sides of our inequality by three without affecting those absolute value signs.
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And we find that the absolute value of π₯ must be less than one-third.
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So the series converges if the absolute value of π₯ is less than one-third.
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And conversely, we can say it will diverge if the absolute value of π₯ is greater than one-third.
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And in fact, weβve found the radius of convergence of our power series.
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Itβs π
equals one-third.
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Now another way of writing the absolute value of π₯ being less than one-third is saying that π₯ must be greater than negative one-third and less than a third.
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And so weβve determined an interval of convergence.
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We certainly know that the power series converges for these values of π₯.
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But we donβt know what happens at the end points of our interval.
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Weβre going to clear some space and substitute π₯ equals negative one-third and π₯ equals one-third into our original power series.
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And weβll see if these series converge or diverge using an alternative test.
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Weβll begin by letting π₯ be equal to negative one-third.
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Then our series is the sum from π equals zero to β of three times negative one-third to the πth power over π plus five.
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Which simplifies to the sum from π equals zero to β of negative one to the πth power over π plus five.
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And you might recognise this.
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Itβs an alternating series.
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This negative one to the πth power indicates so.
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So weβre going to use the alternating series test to establish whether this converges or diverges.
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This series is to be used when we have a series the sum of π π, where π π is negative one to the πth power times π π or negative one to the power of π plus one times π π.
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And here π π must be greater than or equal to zero for all π.
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Then if the limit as π approaches β of π π equals zero and π π is a decreasing sequence, then we can say our series the sum of π π is convergent.
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Letβs rewrite our series as the sum from π equals zero to β of negative one to the πth power times π plus five to the power of negative one.
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We now see itβs of the form given, where π is equal to π plus five to the power of negative one.
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Letβs check the limit as π approaches β of this π π is indeed zero.
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Well, we know itβs equal to the limit as π approaches β of one over π plus five.
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And as π π gets larger, π plus five, the denominator of the fraction, gets larger and larger.
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So one over π plus five gets smaller and smaller and eventually approaches zero.
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So the limit of π π as π approaches β is indeed zero.
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Next, we need to check whether π π is a decreasing sequence.
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To do so, letβs differentiate the entire function with respect to π.
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Using the general power rule, we see thatβs negative π plus five to the power of negative two, which is negative one over π plus five all squared.
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For all values of π between zero and β, the denominator of our fraction is always positive.
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So negative one over a positive number gives us a negative number, which therefore means that we do indeed have a decreasing sequence.
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Weβve satisfied all the criteria for our series.
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And we can therefore say that the series when π₯ is equal to negative one-third converges.
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Weβre now going to clear some space and test π₯ equals a third.
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This time, our series is the sum from π equals zero to β of three times a third to the πth power over π plus five, which is one to the πth power over π plus five, which is of course simply one over π plus five.
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Now this is actually an example of a general harmonic series.
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And we of course know that these diverge.
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Our series therefore diverges for values of π₯ greater than or equal to negative one-third and less than one-third.
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And our interval is shown.