Video: Determining If a Given Series Converges or Diverges

Determine whether the series βˆ‘_(𝑛 = 1)^(∞) (βˆ’1)^(𝑛 + 1) (5^(𝑛)/𝑛!) converges or diverges.

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Video Transcript

Determine whether the series the sum from 𝑛 equals one to ∞ of negative one raised to the power of 𝑛 plus one multiplied by five to the 𝑛th power divided by 𝑛 factorial converges or diverges.

The question wants us to determine the convergence or divergence of the series. We can also notice that five to the 𝑛th power is positive and 𝑛 factorial is positive. So five to the 𝑛th power divided by 𝑛 factorial is positive. So our summand is negative one to the power of 𝑛 plus one being multiplied by some positive sequence. So this is an alternating series.

Since the series given to us in the question is an alternating series, it makes sense to attempt to try using the alternating series test. Which tells us if the sequence π‘Ž 𝑛 is eventually nonnegative, the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero, and π‘Ž 𝑛 is a decreasing sequence. Then we can conclude that the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one multiplied by π‘Ž 𝑛 converges.

So if we set π‘Ž 𝑛 to be equal to five to the 𝑛th power divided by 𝑛 factorial, and we can show that the prerequisites of our alternating series tests are true, then we can conclude that our alternating series given to us in the question converges. We know that five to the 𝑛th power is positive and 𝑛 factorial is positive. So five to the 𝑛th power divided by 𝑛 factorial is going to be a positive number. So π‘Ž 𝑛 is always positive.

To determine the limit as 𝑛 approaches ∞ of five to the 𝑛th power divided by 𝑛 factorial, let’s look at one term of π‘Ž 𝑛. We can expand our numerator of five to the 𝑛th power to be five multiplied by five multiplied by five. And we just do this 𝑛 times. And we can expand the denominator 𝑛 factorial to be 𝑛 multiplied by 𝑛 minus one multiplied by β€” and we go all the way down to one. Now, we can split our fraction dividing each of the fives in our numerator by one of the elements in our denominator. This gives us five over 𝑛 multiplied by five over 𝑛 minus one multiplied by β€” and we go all the way down to five divided by one.

The next thing we notice is that five divided by six is less than one. So when our denominator is equal to six, we’re multiplying by a number, which is less than one. So let’s let 𝑛 be greater than or equal to six. So that means that five divided by six appears somewhere in our product. Now, when the denominator in our fraction is bigger than six, we have that this fraction must be less than or equal to five divided by six. Since all of the terms in our product to the left of five over six are less than or equal to five over six, we can replace all of these with five over six and change our equality to a less than or equal sign.

Since there are 𝑛 total terms in our product and five of these are not equal to five divided by six, 𝑛 minus five of these are equal to five divided by six. So we can replace these with five divided by six raised to the power of 𝑛 minus five. This gives us that our product is equal to five divided by six to the power of 𝑛 minus five multiplied by five to the fifth power divided by five factorial. And we can evaluate five to the fifth power divided by five factorial to be equal to 625 divided by 24.

We’re now ready to use a particular version of the squeeze theorem, which tells us that if we can eventually bound our sequence π‘Ž 𝑛 between zero and 𝑏 𝑛 and the limit of 𝑏 𝑛 as 𝑛 approaches ∞ is equal to zero. Then we can conclude that the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is also equal to zero. We’re going to set 𝑏 𝑛 to be equal to the upper bound we found earlier. We’ve already shown that π‘Ž 𝑛 is a positive sequence. And we know that when 𝑛 is greater than or equal to six, π‘Ž 𝑛 is less than or equal to 𝑏 𝑛. So our first prerequisite for the squeeze theorem is true.

So let’s calculate the limit as 𝑛 approaches ∞ of our sequence 𝑏 𝑛. We can take the constant coefficient of 625 divided by 24 outside of our limit to get 625 divided by 24 all multiplied by the limit as 𝑛 approaches ∞ of five-sixths to the 𝑛th power. And we know from geometric sequences that the limit as 𝑛 approaches ∞ of π‘Ÿ to the 𝑛th power is equal to zero, if the absolute value of our ratio π‘Ÿ is less than one. So since five-sixths is less than one, our limit as 𝑛 approaches ∞ of five-sixths to the 𝑛th power is equal to zero.

So we have shown that the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is equal to zero. So by the squeeze theorem, we can conclude that the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero, which is another one of our prerequisites for our alternating series test. We can actually change the last prerequisite in our alternating series test, since π‘Ž 𝑛 only needs to be an eventually decreasing sequence.

To determine whether π‘Ž 𝑛 is a decreasing sequence, let’s try to rewrite π‘Ž 𝑛 plus one in terms of π‘Ž 𝑛. We can do this by noticing that five to the power of 𝑛 plus one is equal to five multiplied by five to the 𝑛th power and 𝑛 plus one factorial is equal to 𝑛 plus one multiplied by 𝑛 factorial. This gives us that π‘Ž 𝑛 plus one is equal to five multiplied by five to the 𝑛th power divided by 𝑛 plus one multiplied by 𝑛 factorial, which is equal to five divided by 𝑛 plus one all multiplied by π‘Ž 𝑛.

Since we want a decreasing sequence, we want π‘Ž 𝑛 plus one to be smaller than π‘Ž 𝑛. And since π‘Ž 𝑛 is a positive number, this will be true when five divided by 𝑛 plus one is smaller than one. And we can see that this is true when the denominator is greater than five, so when 𝑛 is greater than or equal to five.

Therefore, we’ve shown the π‘Ž 𝑛 is an eventually decreasing sequence, which was the final prerequisite for our alternating series test. So we can conclude by using the alternating series test that the sum from 𝑛 equals one to ∞ of negative one raised to the power of 𝑛 plus one multiplied by π‘Ž 𝑛 converges. And so we’ve shown that the sum from 𝑛 equals one to ∞ of negative one raised to the power of 𝑛 plus one multiplied by five to the 𝑛th power divided by 𝑛 factorial converges.

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