Hi everyone.
Welcome back to integralcalc.com.
We're going to be doing another example today
of interval of convergence.
The sum that we have today is 5 to the n times
x to the n divided by the quantity n factorial.
As with any interval of convergence problem,
the first thing that we're going to do is,
we're going to substitute n plus 1  everywhere
we see n on our original function, and  we're
going to divide that whole thing by the original
function.
So you can see on the top part of our fraction
that we've substituted everywhere n plus 1
for n and then we divide that whole thing
by our original function.
We're taking the limit as n approaches infinity
always.
So once we set that up, now it's just a matter
of simplifying what's inside these absolute
value brackets.
The way that we do that first is instead of
dividing by this entire fraction, we're going
to multiply the top fraction by the inverse
of the bottom fraction.
So we flip this fraction on its head and we
have this n factorial on the numerator and
the 5 to the n times x to the n in the denominator
and we're multiplying instead of dividing.
Next step on our simplification process, I'm
going to rearrange these fractions.
Since everything is multiplied together, I
could do that and I'm going to put this n
plus 1 factorial and this n plus 1 together
and then this 5 to the n plus 1 times x to
the n plus 1 divide by 5 to the n times x
to the n.
I'm putting those together because it's going
to make it easier for me to simplify as you'll
see next.
So right now, I'm going to try to tackle for
simplification purposes this first fraction
here, this n factorial divided by the quantity
n plus 1 factorial.
And the way that I'm going to do that just
to illustrate how this simplifies.
If we subsitute a random number, in this case
we'll say 4 for n, 4 factorial is 4 x 3 x
2 x 1, which we have in the numerator.
In the denominator, we’ll have 4 plus 1
which will be five so we have 5 factorial
and that would be 5 x 4 x 3 x 2 x 1.
So you can see that no matter what number
you pick for n here, all of these terms in
the numerator are going to cancel with every
term in the denominator except this first
one here.
So, all that's going to remain is 1/5.
That means that everything on the numerator
here is going to cancel and we're just going
to be left with n plus 1, no factorial in
the denominator.
So I'm going to go ahead and change this first
fraction to be 1 divided by n plus 1 which
is the simplification of that fraction there.
Now I'm going to go ahead and tackle this
second fraction here.
And the way that we do that is with the following
method.
Remember that when you have a fraction and
you have something both in the numerator and
the denominator on the same base, in other
words, here we have 5 raised to an exponent
in the numerator and denominator, we also
have n raised to some exponent in the numerator
and n raised to some exponent in the denominator.
So those are like bases.
The 5 and 5, the x and x.
We have the same base just different exponents.
When that's the case, you can subtract the
exponent in the denominator from the exponent
in the numerator.
So in our case, we have n plus 1 in the numerator
and so we're going to subtract from that just
the n in the denominator.
n plus 1 minus n just gives you 1, which means
that everything on our denominator is going
to cancel and we're just going to be left
with 5x in the numerator.
So we've simplified now our two fractions
and now what we're going to do is go ahead
and pull out this 5x.
The reason that we can do that is because
we're taking the limit as n approaches infinity.
So all that matters for what's inside our
absolute value brackets here, is anything
that's related to n so the 5x part, we can
remove.
Because it was inside absolute value brackets,
we have to keep it inside absolute value brackets.
Then we're multiplying the absolute value
of 5x by the limit as n approaches infinity
of this 1 divided by n plus 1.
Now that we've done that, we can go ahead
and plug in infinity for n, because we've
simplified enough that we can just plug in
infinity easily.
And you see that we end up with 1 divided
by infinity plus 1.
In other words, 1 divided by infinity.
The 1 divided by infinity is going to give
us 0 because when we take 1 and we divide
it by a huge, huge, huge, huge, huge number,
that's going to give you a very, very, very,
very, very tiny number close to 0 which means
that the limit as n approaches infinity is
going to become 0.
It’s going to approach 0.
So this whole second part, the limit as n
approaches infinity of this whole function
is all going to become 0.
And we're going to be left with the absolute
value of 5 x times 0 which of course is going
to equal 0.
So normally, these interval of convergence
problems end up being longer.
We would set something like 5x to be less
than 1 and then we would move on with our
interval of convergence.
However, it's important to illustrate and
that's why I wanted to do this example.
Whenever you end up with 0 after you take
this limit here, it means that the interval
of convergence is all real numbers, which
looks like this, our final answer.
The interval of convergence is all real numbers
so everything from negative infinity to positive
infinity.
So whenever you end up with 0 at this interval
of convergence, negative infinity to positive
infinity is going to be our final answer.
Just keep that in mind when you end up with
this in the step.
This is going to be your final answer.
You don't need to go any further to test end-points
or anything like that.
So I hope that helped you, guys and I'll see
you in the next video.
Bye.
