# Determining the Radius and Interval of Convergence for a Power Series

## Key Questions

• Given a real power series ${\sum}_{n = 0}^{+ \infty} {a}_{n} {\left(x - {x}_{0}\right)}^{n}$, the radius of convergence is the quantity r = "sup" \{tilde{r} \in \mathbb{R} : sum_{n=0}^{+infty}a_n tilde{r}^n " converges"\}. Note that $r \ge 0$, because for $t i l \mathrm{de} \left\{r\right\} = 0$ the series ${\sum}_{n = 0}^{+ \infty} {a}_{n} t i l \mathrm{de} {\left\{r\right\}}^{n} = {\sum}_{n = 0}^{+ \infty} {a}_{n} {0}^{n} = 1$ converges (recall that ${0}^{0} = 1$).

This quantity it's a bound to the value taken by $| x - {x}_{0} |$. It's not hard to prove that the given power series will converge for every $x$ such that $| x - {x}_{0} | < r$ and it will not converge if $| x - {x}_{0} | > r$ (the proof is based on the direct comparison test). The convergence of the case $r = | x - {x}_{0} |$ depends on the specific power series.

This means that the interval $\left({x}_{0} - r , {x}_{0} + r\right)$ (the interval of convergence) is the interval of the values of $x$ for which the series converges, and there are no other values of $x$ for which this happens, except for the two endpoints ${x}_{+} = {x}_{0} + r$ and ${x}_{-} = {x}_{0} - r$ for which the convergence has to be tested case-by-case.
The term radius is thereby appropriate, because $r$ describes the radius of an interval centered in ${x}_{0}$.

The definition of radius of convergence can also be extended to complex power series.

• By Ratio Test, we can find the radius of convergence: $R = 1$.

By Ratio Test, in order for ${\sum}_{n = 0}^{\infty} {a}_{n}$ to converge, we need
$\setminus {\lim}_{n \to \infty} | \frac{{a}_{n + 1}}{{a}_{n}} | < 1$.

For the posted power series, ${a}_{n} = {x}^{n}$ and ${a}_{n + 1} = {x}^{n + 1}$.
So, we have
$\setminus {\lim}_{n \to \infty} | \frac{{x}^{n + 1}}{{x}^{n}} | = {\lim}_{n \to \infty} | x | = | x | < 1 = R$

Hence, its radius of convergence is $R = 1$.

• The interval of convergence of a power series is the set of all x-values for which the power series converges.

Let us find the interval of convergence of ${\sum}_{n = 0}^{\infty} \frac{{x}^{n}}{n}$.
By Ratio Test,
lim_{n to infty}|{a_{n+1}}/{a_n}| =lim_{n to infty}|x^{n+1}/{n+1}cdotn/x^n| =|x|lim_{n to infty}n/{n+1}
$= | x | \cdot 1 = | x | < 1 R i g h t a r r o w - 1 < x < 1$,
which means that the power series converges at least on $\left(- 1 , 1\right)$.

Now, we need to check its convergence at the endpoints: $x = - 1$ and $x = 1$.

If $x = - 1$, the power series becomes the alternating harmonic series
${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / n$,
which is convergent. So, $x = 1$ should be included.

If $x = 1$, the power series becomes the harmonic series
${\sum}_{n = 0}^{\infty} \frac{1}{n}$,
which is divergent. So, $x = 1$ should be excluded.

Hence, the interval of convergence is $\left[- 1 , 1\right)$.