Video Transcript
Find the partial sum of the series
the sum from 𝑛 equals one to ∞ of two divided by 𝑛 plus three times 𝑛 plus
four. Is the series convergent or
divergent?
The question wants us to find an
expression for the partial sum of the series. It then wants us to determine if
this series is convergent or divergent. We recall that a partial sum 𝑆 𝑘
is the sum of the first 𝑘 terms of our series. So this gives us that our 𝑘th
partial sum, 𝑆 sub 𝑘, is equal to the sum from 𝑛 equals one to 𝑘 of two divided
by 𝑛 plus three times 𝑛 plus four. And we notice that we can simplify
our summand by using partial fractions.
Since we have two distinct factors
in our denominator, we can write two divided by 𝑛 plus three times 𝑛 plus four as
𝐴 divided by 𝑛 plus three plus 𝐵 divided by 𝑛 plus four for some constants 𝐴
and 𝐵. Next, we multiply both sides of
this equation by the denominator 𝑛 plus three times 𝑛 plus four. This gives us two is equal to 𝐴
times 𝑛 plus four plus 𝐵 times 𝑛 plus three. In fact, this will be true for all
values of 𝑛. We represent this by using an
equivalent sign.
Since this is true for all values
of 𝑛, we can eliminate our variables by using substitution. Substituting 𝑛 is equal to
negative four gives us that two is equal to zero minus 𝐵. Therefore, multiplying both sides
of our equation by negative one gives us that negative two is equal to 𝐵. Similarly, we can eliminate the
variable 𝐵 from our equation by substituting 𝑛 is equal to negative three. Substituting 𝑛 is equal to
negative three and then simplifying gives us that 𝐴 is equal to two.
So we’ve shown by partial fractions
that two divided by 𝑛 plus three times 𝑛 plus four is equal to two divided by 𝑛
plus three minus two divided by 𝑛 plus four. So we can use this to simplify the
summand in our partial sum. So by using partial fractions,
we’ve shown that our 𝑘th partial sum is equal to the sum from 𝑛 equals one to 𝑘
of two divided by 𝑛 plus three minus two divided by 𝑛 plus four.
We can now evaluate this by writing
it out term by term. To get our first term, we
substitute 𝑛 is equal to one. Then we add the second term where
𝑛 is equal to two. And we do this all the way up to
the 𝑘th term, where 𝑛 is equal to 𝑘. Simplifying the denominators in our
first term gives us two over four minus two over five. Simplifying the denominators in our
second term gives us two over five minus two over six. And we see that the negative two
over five in our first term and the two over five in our second term cancel. We would see a similar story in our
third term. The negative two over six cancels
with the positive two over six.
And now we see a pattern. Every time we add a new term, it
cancels with the last part of the previous term. We call a series with this property
a telescopic series. So when we added our final term of
two divided by 𝑘 plus three minus two divided by 𝑘 plus four, the first part
canceled with the last part of the previous term. By simplifying our first term to a
half, we’ve shown that the 𝑘th partial sum is equal to a half minus two divided by
𝑘 plus four.
We can simplify this into one
fraction by using cross-multiplication. This gives us one times 𝑘 plus
four minus two times two all divided by two times 𝑘 plus four. And we see that our numerator
simplifies to just give us 𝑘. So we’ve shown that the 𝑘th
partial sum is equal to 𝑘 divided by two times 𝑘 plus four.
In fact, we don’t have to use the
𝑘th partial sum. We could use the 𝑛th partial sum,
giving us that the 𝑛th partial sum, 𝑆 𝑛, is equal to 𝑛 divided by two times 𝑛
plus four.
The next part of the question is to
determine whether our series is convergent or divergent. And we recall that we call a series
convergent if the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is equal to some
finite value 𝐿. Otherwise, we call the series
divergent.
So to check if our series is
convergent or divergent, we need to check the limit as 𝑛 approaches ∞ of the 𝑛th
partial sum. We’ve shown that this is equal to
the limit as 𝑛 approaches ∞ of 𝑛 divided by two times 𝑛 plus four. Since the factor of one-half is
constant with respect to 𝑛, we can take it outside of our limit.
There are then several different
ways of evaluating this limit. We could use algebraic division to
divide 𝑛 by 𝑛 plus four. Or we could divide by the highest
power of 𝑛 in our numerator and our denominator. The highest power of 𝑛 in our
numerator or our denominator is just 𝑛 itself. So we’ll divide both of these
through by 𝑛. Dividing our numerator of 𝑛 by 𝑛
just gives us one. We can then divide each term in our
denominator by 𝑛. This gives us one plus four divided
by 𝑛.
Now we see that our numerator of
one remains constant as 𝑛 is approaching ∞. In our denominator, the one remains
constant as 𝑛 is approaching ∞. However, four divided by 𝑛
approaches zero. That means this limit is
approaching one, which means we can evaluate the limit as 𝑛 approaches ∞ of our
𝑛th partial sum to be one-half.
Therefore, since the limit of our
𝑛th partial sum approached a finite value of one-half, we can conclude that our
series is convergent. Therefore, by finding the partial
sum of the series the sum from 𝑛 equals one to ∞ of two divided by 𝑛 plus three
times 𝑛 plus four. We’ve shown not only is this series
convergent, but we’ve shown that it converges to the value of one-half.
