So in this example we need to find the
interval of convergence of these power
series. Whenever we need to find the
interval of convergence of power series
firstly look at the power series
composed of the absolute values of each
term. So we look at this series and to
find the interval of convergence we
apply the ratio test. So we denote by a_k
the term and this series so this x-2^k
the absolute value and
over k^3. We can of
course write it as the absolute
value of k we can of course rewrite this
as the absolute value of x-2^k/k^3.
Now we will apply the ratio test.
The ratio tells us that
we should take the number r which is
the limit when k goes to infinity of a_k+1/a_k.
So in this case we will
have limit k goes to infinity x-2^k+1/a_k. So in this case we will have limit k goes to infinity
x-2^k+1/k+1^3
and we have here a_k so
this x-2 in the bottom and k^3 to
on the top of the fraction.
So we will rewrite this limit in the
following form.
So this is the limit k goes to infinity the
absolute value of x-2^k+1
over the absolute value of x-2^k
and multiplied by
the other fraction which is k^3/k+1^3.
Of course we can write it as limit k goes
to infinity the absolute value of x-2
x are the same fraction in here.
So with respect to k this term is just a
constant. So we can put this out of the
limit, so it will be x-2
multiplied by the limit k goes to
infinity with the same fraction in here.
Of course when k goes to infinity this
limit is just 1 so we have that this
equal the absolute value of x-2. So
we have that r and here equals to x-2
by the ratio test the series,
the original series in here, will
converge when r is less than 1.
Rewriting this in terms of x we will
have that the absolute value of x-2
is less than 1. And now we can of
course write it as x-2 less than 1
but greater than -1 and this is
equivalent to x less than 3 and bigger
than 1. So the ratio test tells us
that for x in the interval 1 to 3 this
series converge absolutely.
So we have found the interval of
convergence of these power series. Now we
need to discuss the behavior of these
power series at the endpoints of this
interval. We will take firstly the end
point x = 3 substituting this
value we will have the series 1^k/k^3. 1^k
is of course just 1, so this is
just the series 1/k^3
this series converge by the convergence
test for p series. This one converges we
will take now another and point x = 1 and we will get the series -1^k/k^3
when we take the absolute values of each
term we will have the series the
absolute value in here and it's just
equal again to the series 1/k^3 again this one converges.
Therefore for both endpoints this power
series converges, and therefore we can
conclude that the series the original
series converges for all
x in the closed interval 1 to 3.
you
