In this example, we're asking to
determine if the given infinite series converges
or diverges, and if it does converge,
find the sum of the series.
This is an example of a telescoping series
where a telescoping series fits the form of the series here
where if we regroup the terms, we'll notice that many
of the terms are opposites,
and therefore they'll simplify to zero.
So our goal here will be to find a simplified formula
for the partial sum S sub n
then if the limit as n approaches infinity
of the partial sum equals L, the finite value,
then the series converges, and it converges to L.
The series will also only converge if and only if
a sub n approaches finite number as n approaches infinity.
So going back to our example, we'll begin
by expanding the series to find the partial sum S sub n.
So notice how the first group of terms in this series
would be when k is equal to two.
So we'd have 1/2 minus 1/4.
Next, k would be three, so we'd have 1/3 minus 1/5.
Next, k is four, so we'd have 1/4 minus 1/6.
Plus, when k is five, we have 1/5 minus 1/7.
And, when k is six, we have 1/6 minus 1/8.
Next we'd have, when k is seven, we'd have 1/7 minus 1/9.
Of course this continues, and the last term
would be k is equal to n.
Let's see if we can find the last two groups of terms.
Again, the last term would be when k is equal to n,
which would be one divided by n minus one divided by
the quantity n plus two, which means the previous
group of terms here would be when k is equal
to n minus one, not n.
So when k is n minus one, notice how the first
fraction is one divided by the quantity n minus one,
and the second fraction would be one divided by the quantity
n minus one plus two, which would be n plus one.
Now let's begin simplifying and see if you
can recognize the pattern to find a formula for S sub n.
So we're looking for terms that are opposites.
Notice here we have minus 1/4 plus 1/4.
Then we also have minus 1/5 plus 1/5.
Now let's pause here and say for example
if we wanted to find the sum of the first
one, two, three, four groups of terms,
notice how we'd have 1/2 plus 1/3
and then minus 1/6 minus 1/7.
Notice how the 1/2 and 1/3 will never simplify out
because these fractions are getting smaller and smaller.
So to find the sum of the first four groups of terms,
1/2 plus 1/3 and the two fractions remaining
are the second fractions from these last two groups.
Now let's continue simplifying
until we recognize the pattern.
So next, notice how we have minus 1/6 plus 1/6.
So if we wanted to find the sum of the first five
groups of terms, again notice how we have
the 1/2 plus 1/3 again,
and the only remaining two fractions are these
last two fractions from the last two groups.
Here and here.
Let's go ahead and simplify again.
The next two fractions to simplify out
would be minus 1/7 plus 1/7.
So we wanted to find the sum
of the first six groups of terms,
notice how we have 1/2 plus 1/3
and then minus 1/8 minus 1/9.
So again we had the 1/2 and 1/3,
the only remaining two fractions are again
the last two fractions from the last two groups.
So that's our pattern.
If we continue simplifying through the nth term,
the only two fractions remaining would be
these last two fractions from the last two groups.
All the other fractions would simplify out.
Which means a simplified formula for the partial sum S sub n
would be equal to 1/2 plus 1/3,
and then we have minus one divided by the quantity
n plus one minus one divided by the quantity n plus two.
So now we'll take the limit as n approaches infinity
of the partial sum S sub n
and notice as n approaches infinity,
for these last two fractions,
denominators increase without bound,
and therefore the value of these fractions
both approach zero.
And therefore the limit of the partial sum
is equal to 1/2 plus 1/3.
So if we obtain a common denominator of six.
Notice how we have 3/6 plus 2/6, which equals 5/6.
So because this limit exists, we know the series converges,
and the sum of the series is 5/6.
I hope you found this helpful.