Video: Determine the Convergence or Divergence of a Series by Using the Ratio Test

Consider the series βˆ‘_(𝑛 = 1)^(∞) ((3^𝑛)/𝑛!). Determine whether the series converges or diverges.

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Video Transcript

Consider the series the sum from 𝑛 equals one to ∞ of three to the 𝑛th power divided by 𝑛 factorial. Determine whether the series converges or diverges.

The question wants us to determine the convergence or divergence of the series given to us in the question. One thing we might want to try is the ratio test, which tells us that if the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is less than one, then the sum from 𝑛 equals one to ∞ of the sequence π‘Ž 𝑛 converges. And if the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is greater than one, then the sum from 𝑛 equals one to ∞ of the sequence π‘Ž 𝑛 diverges. So this gives us a test which checks both the convergence and divergence, simultaneously.

It’s also worth noting that it’s useful to think of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛 as π‘Ž 𝑛 plus one multiplied by the reciprocal of π‘Ž 𝑛. So let’s set our sequence π‘Ž 𝑛 to be equal to three to the 𝑛th power divided by 𝑛 factorial. And let’s calculate the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms. First, we have that π‘Ž 𝑛 plus one is equal to three to the power of 𝑛 plus one divided by 𝑛 plus one factorial. And then, we multiply this by the reciprocal of π‘Ž 𝑛, which is 𝑛 factorial divided by three to the 𝑛th power. We have 𝑛 plus one factors of three in our numerator and 𝑛 factors of three in our denominator. So we can cancel this out to just get one factor of three in our numerator.

Next, we can rewrite 𝑛 plus one factorial to be equal to 𝑛 plus one multiplied by 𝑛 factorial. So if we cancel the shared factor of 𝑛 factorial in both the numerator and the denominator, we just have a factor of 𝑛 plus one in our denominator. This gives us the limit as 𝑛 approaches ∞ of the absolute value of three divided by 𝑛 plus one. Three is just a constant, so we can take the three outside of our limit where we keep the absolute value sign. This gives us the absolute value of three multiplied by the limit as 𝑛 approaches ∞ of the absolute value of one divided by 𝑛 plus one.

Finally, we know that the limit as 𝑛 approaches ∞ of one divided by 𝑛 plus one is equal to zero. So the limit of the absolute value of the ratio of our successive terms is equal to zero. Since this value of zero is less than one, we’ve shown that the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms is less than one. So we can conclude by using the ratio test that the sum from 𝑛 equals one to ∞ of three to the 𝑛th power divided by 𝑛 factorial must converge.

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