Hey guys, in this video I'm going to do this short real analysis proof. It says
prove that the infinite sum of 7 to the n over 2 to the 3n is a convergent
series and find its sum. We're given a series or infinite sum and we're also
told that it's convergent but we have to prove that and once we're done proving
that, we also have to find the sum of this infinite series. Let's go ahead
and get started. When you're in a real analysis class you're going to want to
identify what type of infinite series you're given because it's going to
probably fall into one of a couple basic categories. I'm going to go over two
broad categories of infinite series and how you can prove--by identifying them
and defining them as a certain type of series--whether they're converging or
diverging. One of those is going to be the p series and this takes on the form
of the infinite sum from 1 to infinity of 1 over n to the p. So you have n that
goes from 1 to infinity and then you're going to have n exponentially be taken
to some number p. This p series is going to converge by definition if p is
greater than 1. If you have a series that looks something like this, the p
series converges by definition if p is greater than 1. If you have the
infinite sum of 1 to infinity of 1 over n to the 3 for example, this would
converge. If you have 1 over n to the 2 that would converge, etc. This type of
broad category of series is called a p series and the second example is going
to be a geometric series, so an infinite geometric series, and this example in
this problem is actually a geometric series. A geometric series has the
following form: a geometric series is broadly defined as the sum from 1 to
infinity of a times r to the n, where a is basically a coefficient. It's going to
be a number. R is a number or a ratio of numbers to an index n.
And this general form is a geometric series--an infinite geometric series--and
it converges by definition if the absolute value of r is less than 1. So r
will usually be a ratio and if that ratio, the absolute value of it, is less
than 1 then it converges by definition. An infinite geometric series
converges by definition if the absolute value of the ratio is less than 1. So if
we go back to our problem, our original series that we're given is finding the
infinite sum from n is equal to 1 to infinity of 7 to the n over 2 to the 3n.
We can rewrite this and prove its convergence as follows. Let's take 7
to the n over 2 to 3n. We can rewrite that as 7 to the n over 2 to the 3 is
equal to 8. So we can rewrite that as 8 to the n and then you have 7 to the n in
the numerator and 8 to the n in the denominator. So we can rewrite that as 7
over 8 to the n. And then the n if you distribute it would go both to the
numerator and the denominator. We can rewrite our original geometric infinite
sum as the sum from n is equal to 1 to infinity of 7 over 8 to the n. So in this
case after we do some simple simplification we see that we have the
form of a times r to the n. In this case we don't really have a coefficient a. We
can just think that a is equal to 1 and our r is equal to 7 over 8 and then n is
our index. For the proof I'm going to say let r equal to 7 over 8. Then the
absolute value of r is less than 1 because the absolute value of 7 over 8
is just 7 over 8, which is less than 1, and then I can write that the sum from n
is equal to 1 to infinity of 7 to the n over 2 to the 3n is equal to the sum of
n is equal to 1 to infinity of 7 over 8 n and this is by definition a convergent
geometric series. So the proof--I'm proving this convergence just by
identifying this as a geometric series and knowing the definition that
a geometric series always converges when r is less than 1--the absolute value of r
is less than 1. Now we have done the first half of the problem, which is
prove that the infinite series converges and the second part of the
problem we have to find the sum. That also has a formula and the formula is as
follows: for an infinite geometric series the sum of that is going to have the
formula a times r over 1 minus r and this is for the infinite series where
the index starts at 1 and then goes to infinity, so you just want to make sure
that you have the correct index. You can also calculate this if n starts at 0 but
the formula is going to be a little bit different. In this case this is the
formula for the type of infinite series that we have. Again like we said
before in our case we don't really have a constant coefficient, it's just going
to be 1. So 1 times anything is itself and then we have that r is equal to 7
over 8 and then in the denominator we're going to have 1 minus 7 over 8. So that's
going to be 7 over 8, over 1 over 8, which is equal to 7. So for this problem we
proved that the infinite series is convergent because we identified it as a
infinite geometric series where the absolute value of r is less than 1 and
then we use the formula a times r over 1 minus r to get the sum and in that case
that gave us that the sum is equal to 7 and we are done with this problem~
*chiptune music*