# What is the interval of convergence of sum ((-1)^n * x^(n+8) ?

Feb 25, 2016

The radius of convergence $R$ of a power series ${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$ is given by $R = \frac{1}{\lim {\text{sup}}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid}}$.

In that case, $R = 1$.

#### Explanation:

Let's write ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {x}^{n + 8}$ in the form of ${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$.

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {x}^{n + 8} = {\sum}_{m = 0}^{\infty} {a}_{m} {x}^{m}$,

where ${a}_{m} = 0 \mathmr{if} m < 8 \mathmr{and} {a}_{m} = {\left(- 1\right)}^{m - 8} \mathmr{if} m \ge 8$.

Now you can compute $R = \frac{1}{\lim {\text{sup}}_{m \to \infty} \sqrt[m]{\left\mid {a}_{m} \right\mid}}$.

Yet, $\left\mid {a}_{m} \right\mid$ can only take $0$ or $1$ as a value. But when $m$ tends to infinity (so when $m$ is much larger than $8$), abs(a_m) = 1 = root(m)(abs(a_m) so $\lim {\text{sup}}_{m \to \infty} \sqrt[m]{\left\mid {a}_{m} \right\mid} = 1$.

Therefore, $R = \frac{1}{1} = 1$.

You now know that ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {x}^{n + 8}$ converges for x in ]-1, 1[.

What about $x = - 1$ or $x = 1$ ?

If $x = - 1$ :
${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {\left(- 1\right)}^{n + 8} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{2 n + 8} = {\sum}_{n = 0}^{\infty} 1 = + \infty$

If $x = 1$ :
${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {\left(1\right)}^{n + 8} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n}$ is not defined.

So ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \cdot {x}^{n + 8}$ converges for x in ]-1, 1[.