Question

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

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

Answer 1

`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`

Answer 2

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.