- [Teacher] So in this video
we're just looking at the
convergent or divergent
of the couple of series.
So the first one here we
have 3 to the K minus 1
over 7 to the K.
And what I'm gonna do before
I go very far with this one,
is re-write it with using
properties of exponents,
so I can re-write this as 3 to the K times
3 to the minus 1
over 7 to the K .
And the 3 to the minus 1,
I'm gonna go ahead and pull
that out of the series,
so 3 to the minus 1 it
doesn't depend on K .
And what's in there, 3 to
the K over 7 to the K ,
I'm gonna write this to the same exponent.
Now these are going from 2 to infinity ,
and notice what I have here,
I have a geometric series,
with common ratio less than 1 ,
so this is going to be
a convergent series.
Now,
so this we know is convergent,
we're even gonna find were converges to,
so we have 3 to the negative 1,
which is one third, and
therefore geometric series
it's the first term over
one minus a common ratio.
Now the first term is 3 7th to the second,
because we told the starter
sequent, our series excuse me,
at 2 not 1, so be careful
about where you start,
when you're doing some
of the geometric series.
So, we have, 1 third,
this is going to be 9 over 49 .
So, 9 over 49, this here ,
this only 7 7th minus 3 7th,
is 4 7th, but we're flipping
and multiplying by 7 over 4.
We're gonna be able to do a
little bit of canceling here,
so that's gonna go to 1
, that's gonna go to 7 .
And so we can say that this converges,
oh we could do a little
bit more excuse me.
That goes to 3, this
converges to 3 over 28 .
So, not all we do we can say converges,
since it's geometric
we can very quickly say
what it converges to .
Let's look at our second one,
our second one is a
little tricky at first.
We have 1 over K to the K ,
so this is not a real
direct root to this one,
but here is what I notice.
We're looking at K values, looking at K .
Greater than or equal to 2 .
And we know K is greater
than or equal to 2 for,
K is greater than or equal to
2, seems a little read on it,
but this is the part I wanna work with ,
this is the part that
is gonna be true for.
Well, if that's the case then,
1 over K , is less than
or equal to 1 over 2 ,
for K greater than or equal to 2 ,
so I just flipped both of these,
which makes foot my inequality .
Well then ,
one over K to the K
is less than or equal
to one half of the K .
And I, one thing we can
do is re-write this.
The summation, K equals 2 , to infinity,
a 1 over K , to the K .
Now, what do we know about
the series from K equals 2
to infinity.
Of one half to the K .
This is a geometric series with
a common ration less than 1.
So we know this converges
because it's geometric.
We can find what it converges to,
but it's not important for
what we doing right now.
So, that means with
what we did right here,
this is very important,
this guy,
converges by the comparison
task, because we showed
it was less than something
that we knew to the convergent.
