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Consider the series the sum from π equals zero to β of negative two to the πth power over π times three to the power of π plus one.
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Calculate the limit as π approaches β of the absolute value of π sub π plus one over π sub π.
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Hence, determine whether the series converges or diverges.
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Letβs look back at our original series.
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We know the general form is the sum of π π.
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So, weβre going to let π π be equal to the bit inside the sum.
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Itβs negative two to the power of π over π times three to the power of π plus one.
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We should now be able to see that in order to calculate the limit as π approaches β of the absolute value of π sub π plus one over π sub π, we need to work out π sub π plus one.
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To achieve this, we replace π with π plus one.
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So, the numerator becomes negative two to the power of π plus one.
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And then, the denominator becomes π plus one times three to the power of π plus one plus one or three to the power of π plus two.
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We need to calculate the limit as π approaches β of the absolute value of the quotient of these.
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Now, if we were to write these in fractional form, it might get a bit confusing.
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So instead, we write it as the limit as π approaches β of the absolute value of negative two to the power of π plus one over π plus one times three to the power of π plus two divided by negative two to the πth power over π times three to the power of π plus one.
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Then, we recall that to divide by a fraction, we simply multiply by the reciprocal of that fraction.
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And so, our limit becomes the absolute value of negative two to the power of π plus one over π plus one times three to the power of π plus two times π times three to the power of π plus one over negative two to the πth power.
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We also know that to divide two numbers whose base is equal, we simply subtract their exponents.
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So, negative two to the power of π plus one over negative two to the power of π is negative two to the power of π plus one minus π, which is negative two to the power of one or simply negative two.
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Similarly, three to the power of π plus one divided by three to the power of π plus two is three to the power of π plus one minus π plus two, which is three to the power of negative one or one-third.
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So, weβre simply left with three on the denominator.
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We can now rewrite our limit in a number of ways.
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Iβve chosen to separate negative two-thirds and π over π plus one.
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So, we want to calculate the limit as π approaches β of the absolute value of negative two-thirds times π over π plus one.
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One of the laws of limits says we can take out any constant factors.
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Now, here weβre going to take out the constant factor of negative two-thirds.
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But of course, since this is the absolute value, we take out the absolute value of negative two-thirds, which is simply two-thirds.
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So, letβs find the limit as π approaches β of the absolute value of π over π plus one.
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And if we were to apply direct substitution right now, weβd get β over β, which is of indeterminate form.
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So instead, we perform some manipulation.
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We divide both the numerator and the denominator of our fraction by the highest power of π on our denominator.
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So, weβre going to divide both the numerator and the denominator by π.
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Letβs clear some space.
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The fraction becomes π over π over π over π plus one over π.
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But of course, π over π is equal to one.
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And we need to calculate two-thirds of the limit as π approaches β of the absolute value of one over one plus one over π.
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Now, as π grows larger, one over π grows smaller.
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In other words, as π approaches β, one over π approaches zero.
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And so, the limit as π approaches β of the absolute value of one over one plus one over π becomes the absolute value of one over one, which is simply one.
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Two-thirds times one is two-thirds.
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And so, weβve calculated our limit.
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The limit as π approaches β of the absolute value of π sub π plus one over π sub π is two-thirds.
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So, how do we use this to establish whether the series converges or diverges?
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Well, letβs say the limit as π approaches β of the absolute value of π sub π plus one over π sub π is equal to πΏ, the constant πΏ.
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The ratio test tells us that if πΏ is less than one, the series is absolutely convergent and hence converges.
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If itβs greater than one, the series diverges.
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And if itβs equal to one, our test is inconclusive.
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Well, our value for πΏ was two-thirds; this is quite clearly less than one.
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And so, by the ratio test, our series converges.