Video: Solving a Separable First Order Differential Equation Involving Properties of Exponents

Determine whether the series βˆ‘_(𝑛 = 1)^(∞) (βˆ’1)^(𝑛 + 1) ((5/(𝑛 + 1)) βˆ’ (5/(𝑛 + 2))) converges or diverges.

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Video Transcript

Determine whether the series which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one multiplied by five over 𝑛 plus one minus five over 𝑛 plus two converges or diverges.

Now, looking at our series, we can see that it looks like an alternating series. This is because of the negative one to the 𝑛 plus one term. Now, an alternating series can be of the form as shown here. However, we’re required to that π‘Ž 𝑛 must be greater than zero. In our case, π‘Ž 𝑛 is equal to five over 𝑛 plus one minus five over 𝑛 plus two. We can check whether this is greater than zero by combining the two fractions. We multiply the fraction on the left by 𝑛 plus two over 𝑛 plus two and multiply the fraction on the right by 𝑛 plus one over 𝑛 plus one. After doing this, we’ll have found a common denominator of 𝑛 plus one multiplied by 𝑛 plus two. Then, we can combine the two fractions. This will give five lots of 𝑛 plus two minus five lots of 𝑛 plus one over 𝑛 plus one multiplied by 𝑛 plus two.

Now, we can multiply through the numerator. And for the final step, we can simplify this numerator. What we’re left with is that π‘Ž 𝑛 is equal to five over 𝑛 plus one multiplied by 𝑛 plus two. Now, 𝑛 can be any integer between one and ∞. Therefore, 𝑛 is always positive. And if 𝑛 is always positive, this means that our π‘Ž 𝑛 must also always be positive. Now, we have confirmed that our series is in fact an alternating Series. We can try and use the alternating series test, which tells us that an alternating series converges if the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to zero and π‘Ž 𝑛 is a decreasing sequence.

Let’s start by finding the limit as 𝑛 tends to ∞ of π‘Ž 𝑛. We have that the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to the limit as 𝑛 tends to ∞ of five over 𝑛 plus one multiplied by 𝑛 plus two. Now, all of the 𝑛s in our limit are on the denominator of our fraction. And all of these 𝑛s are positive. Therefore, as 𝑛 gets larger and larger and larger, the denominator of this fraction will also get larger. This means that the whole fraction itself will tend towards zero. Hence, we can say that this limit must be equal to zero. Therefore, we’ve satisfied the first condition of the alternating series test.

Next, we need to check whether the sequence π‘Ž 𝑛 is decreasing. If the sequence is decreasing, then π‘Ž 𝑛 must be greater than π‘Ž 𝑛 plus one, since each term will be smaller than the previous one. We can rearrange this inequality. And here, what we have to check is that π‘Ž 𝑛 minus π‘Ž 𝑛 plus one is greater than zero. For π‘Ž 𝑛, we have five over 𝑛 plus one times 𝑛 plus two. Then, for π‘Ž 𝑛 plus one, we have five over 𝑛 plus two times 𝑛 plus three. We can find the difference here by multiplying the fraction on the left by 𝑛 plus three over 𝑛 plus three and the fraction on the right by 𝑛 plus one over 𝑛 plus one.

After combining the two resulting fractions, what we’re left with is five lots of 𝑛 plus three minus five lots of 𝑛 plus one over 𝑛 plus one times 𝑛 plus two times 𝑛 plus three. Simplifying this fraction, what we’re left with is 10 over 𝑛 plus one times 𝑛 plus two times 𝑛 plus three. Since 𝑛 must be between one and ∞, this tells us that 𝑛 itself must be greater than or equal to one. For any 𝑛 value, which is greater than or equal to one, our fraction here, which is π‘Ž 𝑛 minus π‘Ž 𝑛 plus one, must be greater than zero. This is because all of the terms in this fraction will be positive. So we have satisfied the condition required for π‘Ž 𝑛 to be a decreasing sequence. And therefore, the second condition for the alternating series test is also satisfied.

From this, we can come to the conclusion that the series which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one multiplied by five over 𝑛 plus one minus five over 𝑛 plus two converges.

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