Find the values of x for which the following series is convergent?

\sum_(n=0)^(\infty)(2x-3)^n

2 Answers
Apr 8, 2018

1<x<2

Explanation:

When trying to determine the radius and/or interval of convergence of power series such as these, it is best to use the Ratio Test, which tells us for a series suma_n, we let

L=lim_(n->oo)|a_(n+1)/a_n|.

If L<1 the series is absolutely convergent (and hence convergent)

If L>1, the series diverges.

If L=1, the Ratio Test is inconclusive.

For Power Series, however, three cases are possible

a. The power series converges for all real numbers; its interval of convergence is (-oo, oo)
b. The power series converges for some number x=a; its radius of convergence is zero.
c. The most frequent case, the power series converges for |x-a|<R with an interval of convergence of a-R<x<a+R where we must test the endpoints to see what happens with them.

So, here,

a_n=(2x-3)^n

a_(n+1)=(2x-3)^(n+1)=(2x-3)(2x-3)^n

So, apply the Ratio Test:

lim_(n->oo)|((cancel((2x-3)^n)(2x-3))/cancel((2x-3)^n))|

|2x-3|lim_(n->oo)1=|2x-3|

So, if |2x-3|<1, the series converges. But we need this in the form |x-a|<R:

|2(x-3/2)|<1

2|x-3/2|<1

|x-3/2|<1/2 results in convergence. The radius of convergence is R=1/2.

Now, let's determine the interval:

-1/2<x-3/2<1/2

-1/2+3/2<x<1/2+3/2

1<x<2

We need to plug x=1, x=2 into the original series to see if we have convergence or divergence at these endpoints.

x=1: sum_(n=0)^oo(2(1)-3)^n=sum_(n=0)^oo(-1)^n diverges, the summand has no limit and certainly doesn't go to zero, it just alternates signs.

x=2: sum_(n=0)^oo(4-3)^n=sum_(n=0)^oo1 diverges as well by the Divergence Test, lim_(n->oo)a_n=lim_(n->oo)1=1 ne 0

Therefore, the series converges for 1<x<2

Apr 8, 2018

We can use the ratio test which says that if we have a series
sum_(n=0)^ooa_n

it is definitely convergent if:
lim_(n->oo)|a_(n+1)/a_n|<1

In our case, a_n=(2x-3)^n, so we check the limit:
lim_(n->oo)|(2x-3)^(n+1)/(2x-3)^n|=lim_(n->oo)|((2x-3)cancel((2x-3)^n))/cancel((2x-3)^n)|=

=lim_(n->oo)|2x-3|=2x-3

So, we need to check when |2x-3| is less than 1:

I made a mistake here, but the above answer has the same method and a correct answer, so just have a look at that instead.