Video: Determine Whether an Alternating 𝑝-Series Converges or Diverges

Determine whether the series βˆ‘_(𝑛 = 1)^∞ cos (π‘›πœ‹)/𝑛³ converges or diverges.

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Video Transcript

Determine whether the series the sum from 𝑛 equals one to ∞ of the cos of 𝑛 multiplied by πœ‹ divided by 𝑛 cubed converges or diverges.

The question gives us a series and asked us to determine whether this series is convergent or divergent. We have seen lots of different rules and tests for checking the divergence and convergence of different series. Sometimes, it can be difficult to work out which rule to use. To help us get an idea of which rule we should use, let’s write out the first four terms of our sum. That is the sum from 𝑛 equals one to four of the cos of 𝑛 multiplied by πœ‹ divided by 𝑛 cubed.

We get a first term of the cos of one multiplied by πœ‹ divided by one cubed. Our second term is the cos of two multiplied by πœ‹ divided by two cubed. And we can then do the same to get our third and fourth terms. We can then start evaluating the numerators. The cos of πœ‹ is equal to negative one and the cos of two πœ‹ is equal to one. Similarly, the cos of three πœ‹ is equal to negative one and the cos of four πœ‹ is equal to one. This gives us negative one over one cubed plus one over two cubed minus one over three cubed plus one over four cubed.

What we can see is that each term seems to be switching in sign. That means this appears to be an alternating series. We can actually show that this is an alternating series. We know that the cos of an even multiple of πœ‹ is equal to one, and the cos of an odd multiple of πœ‹ is equal to negative one. What this means is that we can write the cos of 𝑛 multiplied by πœ‹ as negative one to the power of 𝑛. Now, we recall from the alternating series test if the sequence π‘Ž 𝑛 is equal to negative one to the power of 𝑛 multiplied by 𝑏 𝑛, where 𝑏 𝑛 is greater than or equal to zero for 𝑛 greater than or equal to one and the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is equal to zero and also that 𝑏 𝑛 is a decreasing sequence. Then, we can conclude that the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is convergent.

What this means is that if we set our sequence π‘Ž 𝑛 to be equal to the cos of π‘›πœ‹ divided by 𝑛 cubed which we just showed is equal to negative one to the power of 𝑛 over 𝑛 cubed and we set 𝑏 𝑛 to be equal to one over 𝑛 cubed. Then, we have that π‘Ž 𝑛 is equal to negative one to the power of 𝑛 multiplied by 𝑏 𝑛. And we can use these in our alternating series test to attempt to discuss the convergence of the series from 𝑛 equals one to ∞ of π‘Ž 𝑛. To use these in the alternating series test, we have three prerequisites that we must show to be true.

We need that the sequence 𝑏 𝑛 is greater than or equal to zero for 𝑛 greater than or equal to one. We need to show that its limit as 𝑛 approaches ∞ is equal to zero. And we need to show that it is a decreasing sequence. Let’s start by showing that 𝑏 𝑛 is a decreasing sequence. Well, when 𝑛 is greater than or equal to one, 𝑛 plus one is positive and 𝑛 is positive. So 𝑛 plus one is bigger than 𝑛. And we also know that 𝑛 plus one cubed must be bigger than 𝑛 cubed. We can then take the reciprocal of this inequality, giving us that one over 𝑛 plus one cubed is less than one over 𝑛 cubed. Well, we take care to flip the sign of our inequality because we’re taking the reciprocal.

And in particular, what we have shown is that when 𝑛 is greater than or equal to one, 𝑏 𝑛 plus one is less than 𝑏 𝑛. So 𝑏 𝑛 is a decreasing sequence. We can also show that the sequence 𝑏 𝑛 is nonnegative when 𝑛 is greater than or equal to zero. Since 𝑏 𝑛 is just equal to one over 𝑛 cubed, if 𝑛 is positive, 𝑏 𝑛 is positive. This means we’ve shown two of our prerequisites.

The last one we need to show is that limit as 𝑛 tends to ∞ of 𝑏 𝑛 is equal to zero. Well, we know 𝑏 𝑛 is equal to one over 𝑛 cubed. So we can replace the 𝑏 𝑛 in our limit with one over 𝑛 cubed. And we know that this limit is equal to zero since the limit of one over 𝑛 to the power of 𝑝, where 𝑝 is greater than or equal to zero as 𝑛 approaches ∞ is just equal to zero.

We have now shown that all three prerequisites for the alternating series test are true. Therefore, we can use the alternating series test to conclude that the sum from 𝑛 equals one to ∞ of the cos of π‘›πœ‹ divided by 𝑛 cubed must converge.

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