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Using the ratio test, determine whether the series the sum from π equals one to β of two times π to the πth power divided by π factorial is convergent, divergent, or whether the test is inconclusive.
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Weβre given a series, and we need to use the ratio test to determine whether this series is convergent, divergent, or whether the ratio test is inconclusive.
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So letβs start by recalling the ratio test.
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The ratio test tells us, for a series the sum from π equals one to β of ππ, where we set π equal to the limit as π approaches β of the absolute value of ππ plus one divided by ππ, then if this value of π is less than one, then the ratio test tells us that our series must be absolutely convergent.
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However, if this value of π is greater than one, then the ratio test tells us that our series must be divergent.
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Finally, if this value of π is equal to one, then the ratio test is inconclusive.
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So to apply the ratio test, we need to calculate this value of π.
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To do this, we need to calculate the limit as π approaches β of the absolute value of the ratio of successive terms in the series given to us in the question.
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So weβll set our sequence ππ to be the summand of our series.
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Thatβs two times π to the πth power divided by π factorial.
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And we need to calculate π, which is the limit as π approaches β of the absolute value of ππ plus one divided by ππ.
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However, before we evaluate this limit, weβll make one small adjustment.
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Instead of dividing by ππ, weβll multiply by the reciprocal of ππ.
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Weβre now ready to start calculating the value of π.
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We need to substitute π plus one and π to find our values of ππ plus one and ππ.
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This gives us the limit as π approaches β of the absolute value of two times π plus one raised to the power of π plus one divided by π plus one factorial times π factorial divided by two times π to the πth power, where itβs worth pointing out we evaluated the reciprocal of ππ by inverting our fraction.
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Weβre now ready to start simplifying.
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First, in our denominator, we can rewrite π plus one factorial as π plus one multiplied by π factorial.
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This then means we can cancel the shared factor of π factorial in our numerator and our denominator.
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But notice we also have π plus one factors of π plus one in our numerator.
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So we can also cancel one of our factors of π plus one.
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Finally, we can also cancel the shared factor of two in our numerator and our denominator.
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This gives us the limit as π approaches β of the absolute value of π plus one all raised to the πth power divided by π raised to the πth power.
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To simplify this expression, weβre going to need to recall one of our laws of exponents.
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π to the power of π divided by π to the power of π is actually equal to π over π all raised to the power of π.
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And we can see weβre dividing two terms both raised to the πth power.
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This gives us the limit as π approaches β of the absolute value of π plus one all divided by π all raised to the πth power.
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Now, all we need to do to evaluate this limit is to evaluate the division by π.
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Weβll just divide both terms in our numerator by π.
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This gives us the limit as π approaches β of the absolute value of one plus one over π all raised to the πth power.
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And this should now start reminding us of a limit result we know.
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We need to remember the limit as π approaches β of one plus one over π all raised to the πth power is equal to Eulerβs constant π.
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This is almost exactly what we have.
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Weβre just taking the absolute value of this limit.
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But remember, our values of π range from one and then increase.
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So one plus one over π is positive.
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And if one plus one over π is positive for all of our values of π, then we also know one plus one over π all raised to the πth power is positive for all of our values of π.
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So we can remove the absolute value signs.
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So this limit evaluates to give us π.
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We found that our value of π is equal to π.
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And we know π is approximately equal to 2.7.
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This means our value of π is one.
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And that means the ratio test is conclusive.
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In fact, the ratio test tells us that our series must be divergent.
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Therefore, by applying the ratio test to the series the sum from π equals one to β of two times π to the πth power divided by π factorial, we were able to show that this series is divergent.