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Consider the series the sum from π is equal to one to β of π cubed over three π factorial.
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Determine whether the series converges or diverges.
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We can try using the ratio test to determine whether the series converges or diverges.
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The ratio test tells us that for some series, which is the sum from π equals one to β of π π with, πΏ being equal to the limit as π tends to β of the absolute value of π π plus one over π π, that firstly, if πΏ is less than one, then our series converges absolutely.
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Secondly, if πΏ is greater than one, then our series diverges.
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And finally, if πΏ is equal to one, then the test is inconclusive.
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Now, in our case, we can see that π π is equal to π cubed over three π factorial.
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Letβs quickly note that this three π factorial in the denominator is equal to three multiplied by π factorial.
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And so, the factorial symbol affects only the π and not the three.
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Next, we can write down what π π plus one is.
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We simply change each π in π π to π plus one.
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And we see that π π plus one is equal to π plus one cubed over three π plus one factorial.
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Now, weβre ready to find πΏ.
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We have that πΏ is equal to the limit as π tends to β of the absolute value of π π plus one over π π.
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We substitute in π π plus one and one over π π.
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Now, weβre ready to start simplifying.
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First, we can cross cancel the factor of three.
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Next, we know that we can rewrite π plus one factorial as π plus one multiplied by π factorial.
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This will allow us to write our limit as the limit as π tends to β of the absolute value of π plus one cubed multiplied by π factorial over π plus one multiplied by π factorial multiplied by π cubed.
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And so, we can cross out the factor of π factorial from the numerator and denominator.
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We can also cancel a factor of π plus one from the denominator with one of the factors of π plus one from the numerator.
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What weβre left with here is the limit as π tends to β of the absolute value of π plus one squared over π cubed.
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Next, we can distribute the square in the numerator.
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Here, it is clear that our highest power of π in the numerator is π squared and our highest power of π in the denominator is π cubed.
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Since weβre taking the limit as π tends to β and the highest power of π in the denominator is greater than the highest power of π in the numerator, this tells us that this limit will be equal to zero.
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What we have found here is that our value of πΏ for this series is zero and zero is less than one.
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So when we look at our rule for the ratio test, we see that weβve satisfied condition number one.
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And this tells us our solution to the question, which is that the sum from π equals one to β of π cubed over three π factorial converges absolutely.