Video Transcript
Determine whether the series the
sum from 𝑛 equals one to ∞ of the cos of 𝑛 multiplied by 𝜋 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 divided by 𝑛 cubed.
We get a first term of the cos of
one multiplied by 𝜋 divided by one cubed. Our second term is the cos of two
multiplied by 𝜋 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 is equal to negative
one and the cos of two 𝜋 is equal to one. Similarly, the cos of three 𝜋 is
equal to negative one and the cos of four 𝜋 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 is equal to one, and the cos of an odd multiple of 𝜋 is equal to
negative one. What this means is that we can
write the cos of 𝑛 multiplied by 𝜋 as negative one to the power of 𝑛. Now, we recall from the alternating
series test if the sequence 𝑎 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 𝑎 𝑛 is convergent.
What this means is that if we set
our sequence 𝑎 𝑛 to be equal to the cos of 𝑛𝜋 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 𝑎 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 𝑎 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 𝑛𝜋 divided by 𝑛 cubed must converge.
