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

## $\setminus {\sum}_{n = 0}^{\setminus \infty} {\left(2 x - 3\right)}^{n}$

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 $\sum {a}_{n}$, we let

$L = {\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} |$.

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 $\left(- \infty , \infty\right)$
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,

${a}_{n} = {\left(2 x - 3\right)}^{n}$

${a}_{n + 1} = {\left(2 x - 3\right)}^{n + 1} = \left(2 x - 3\right) {\left(2 x - 3\right)}^{n}$

So, apply the Ratio Test:

${\lim}_{n \to \infty} | \left(\frac{\cancel{{\left(2 x - 3\right)}^{n}} \left(2 x - 3\right)}{\cancel{{\left(2 x - 3\right)}^{n}}}\right) |$

$| 2 x - 3 | {\lim}_{n \to \infty} 1 = | 2 x - 3 |$

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

$| 2 \left(x - \frac{3}{2}\right) | < 1$

$2 | x - \frac{3}{2} | < 1$

$| x - \frac{3}{2} | < \frac{1}{2}$ results in convergence. The radius of convergence is $R = \frac{1}{2.}$

Now, let's determine the interval:

$- \frac{1}{2} < x - \frac{3}{2} < \frac{1}{2}$

$- \frac{1}{2} + \frac{3}{2} < x < \frac{1}{2} + \frac{3}{2}$

$1 < x < 2$

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

$x = 1 : {\sum}_{n = 0}^{\infty} {\left(2 \left(1\right) - 3\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n}$ diverges, the summand has no limit and certainly doesn't go to zero, it just alternates signs.

$x = 2 : {\sum}_{n = 0}^{\infty} {\left(4 - 3\right)}^{n} = {\sum}_{n = 0}^{\infty} 1$ diverges as well by the Divergence Test, ${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} 1 = 1 \ne 0$

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
${\sum}_{n = 0}^{\infty} {a}_{n}$

it is definitely convergent if:
${\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} | < 1$

In our case, ${a}_{n} = {\left(2 x - 3\right)}^{n}$, so we check the limit:
${\lim}_{n \to \infty} | {\left(2 x - 3\right)}^{n + 1} / {\left(2 x - 3\right)}^{n} | = {\lim}_{n \to \infty} | \frac{\left(2 x - 3\right) \cancel{{\left(2 x - 3\right)}^{n}}}{\cancel{{\left(2 x - 3\right)}^{n}}} | =$

$= {\lim}_{n \to \infty} | 2 x - 3 | = 2 x - 3$

So, we need to check when $| 2 x - 3 |$ 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.