Real Power Series#

Definition: Real Power Series

A real power series is an expression of the form

\[ \sum_{n\in \mathcal{D}} a_n (x-c)^n, \]

where \((a_n)_{n \in \mathcal{D}}\) is a real sequence and \(c \in \mathbb{R}\).

Intuition

By plugging in a concrete value \(x^\ast \in \mathbb{R}\) for \(x\), one obtains the real series

\[ \sum_{n \in \mathcal{D}} a_n (x^{\ast} - c)^n \]

Convergence#

Definition: Convergence and Divergence of Power Series

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x-c)^n\) be a real power series and let \(z_0 \in \mathbb{R}\).

We say that \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x-c)^n\)

is convergent or converges for \(z_0\) if the resultant real series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is convergent;
is absolutely convergent or converges absolutely for \(z_0\) if the resultant real series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is absolutely convergent;
is divergent or diverges for \(z_0\) if the resultant real series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is divergent;

Theorem: Convergence Intervals

If the real power series \(\sum_{n \in \mathcal{D}} a_n (x-c)^n\) converges for \(w \in \mathbb{R}\), then it also converges for all \(w' \in \mathbb{R}\) with \(|w' - c| \lt |w - c|\).

Proof

Since the series \(\sum_{n \in \mathcal{D}} a_n (w-c)^n\) converges, we know that the limit of \(a_n (w - c)^n\) must be \(0\):

\[ \lim_{n \to \infty} a_n (w-c)^n = 0 \]

Since \(a_n (w - c)^n\) is convergent, it must be bounded, i.e. there exists \(B \in \mathbb{R}_{\gt 0}\) such that

\[ |a_n (w-c)^n| \le B \]

for all \(n \in \mathcal{D}\).

For \(w'\) we have

\[ |a_n (w' - c)^n| = \left|a_n(w' - c)^n \frac{(w-c)^n}{(w-c)^n}\right| = \left|a_n(w - c)^n\right|\left| \frac{w'-c}{w-c}\right|^n \]

Since \(|a_n (w-c)^n| \le B\) for all \(n\in\mathcal{D}\), we know that

\[ |a_n (w' - c)^n| \le B \left| \frac{w'-c}{w-c}\right|^n \]

for all \(n \in \mathcal{D}\). Let \(q = \left| \frac{w'-c}{w-c}\right|\), i.e.

\[ |a_n (w' - c)^n| \le B q^n \]

Since \(|w'-c| \lt |{w-c}|\), we know that \(q \lt 1\). Therefore, \(\sum_{n \in \mathcal{D}} B q^n\) is convergent (\(\sum_{n\in\mathcal{D}}q^n\) is a geometric series). The convergence of \(\sum_{n \in \mathcal{D}} a_n(w'-c)^n\) is thus guaranteed by the majorant convergence criterion.

Theorem: Radius of Convergence

For each real power series \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x - c)^n\), there exists some \(r \in \mathbb{R}_{\ge 0} \cup \{\infty\}\) such that:

If \(r = 0\), then the power series converges only for \(x = c\).
If \(r \in (0; \infty)\), then the power series converges absolutely for all \(x \in \mathbb{R}\) with \(|x - c| \lt r\) and diverges for all \(x \in \mathbb{R}\) with \(|x - c| \gt r\). The behavior for \(x \in \mathbb{R}\) with \(|x| = r\) must be examined separately.
If \(r = \infty\), then the power series converges absolutely for all \(x \in \mathbb{R}\).

Definition: Radius of Convergence

We call \(r\) the radius of convergence.

Proof

TODO

Theorem: Radius of Convergence via Quotient Limit

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x - c)^n\) be a real power series.

If the limit of \(\left|\frac{a_n}{a_{n+1}}\right|\) exists or is \(\infty\), then it is equal to the radius of convergence \(r\):

\[ r = \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| \in \mathbb{R}_{\ge 0} \cup \{\infty\} \]

Proof

TODO

Example: \(\sum_{n = 0}^{\infty} x^n\)

We examine the real power series

\[ \sum_{n = 0}^{\infty} x^n \]

To see that it is indeed a real power series, we rewrite it a bit:

\[ \sum_{n = 0}^{\infty} a_n(x - 0)^n \qquad a_n = 1 \]

We thus have the following:

\[ r = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1 \]

Therefore, \(\sum_{n = 0}^{\infty} x^n\) is absolutely convergent when \(|x| \lt 1\) and divergent when \(|x| \gt 1\). Moreover, we have noticed that \(\sum_{n = 0}^{\infty} x^n\) is a geometric series for each \(x \in \mathbb{R}\). Therefore, we know that it is also divergent when \(|x| = 1\).

Example: \(\sum_{n=1}\frac{1}{n}x^n\)

We examine the following real power series:

\[ \sum_{n=1}\frac{1}{n}x^n \]

To see that it is indeed a real power series, we can rewrite it a bit:

\[ \sum_{n = 1}^{\infty}a_n (x - c)^n \qquad a_n = \frac{1}{n}, c = 0 \]

We therefore have:

\[ r = \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| = \lim_{n \to \infty}\left|\frac{n+1}{n}\right| = 1 \]

For \(x = -1\), we have convergence by the Leibniz convergence criterion. For \(x = 1\), we get the harmonic series which is divergent.

Example: \(\sum_{n = 1}^{\infty} \frac{1}{n^2} x^n\)

We examine the following real power series:

\[ \sum_{n=1}\frac{1}{n^2}x^n \]

To see that it is indeed a real power series, we can rewrite it a bit:

\[ \sum_{n = 1}^{\infty}a_n (x - c)^n \qquad a_n = \frac{1}{n^2}, c = 0 \]

We therefore have:

\[ r = \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| = \lim_{n \to \infty}\left|\frac{(n+1)^2}{n^2}\right| = 1 \]

For \(x = \pm 1\), the power series is absolutely convergent by the majorant convergence criterion, since \(\sum_{n = 1}^{\infty} \frac{1}{n^2}\) is convergent.

Cauchy-Hadamard Theorem

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x - c)^n\) be a real power series.

The limit superior of the sequence \(\sqrt[n]{|a_n|}\) can be used to find the radius of convergence:

If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} = 0\), then \(R = \infty\).
If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} \in (0; \infty)\), then \(R = \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}}\).
If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} = \infty\), then \(R = 0\).

Proof

TODO

Example: \(\sum_{k = 0}^{\infty} x^{k^2}\)

We examine the following real power series:

\[ \sum_{k = 0}^{\infty} x^{k^2} = 1 + x + x^4 + x^9 + x^{16} + \cdots \]

To see that this is indeed a real power series, we need to rewrite it a bit:

\[ \sum_{n = 0}^{\infty} a_n(x-0)^{n} \qquad a_n = \begin{cases}1 & \text{if } n = k^2, k \in \mathbb{N}_0 \\ 0 & \text{otherwise}\end{cases} \]

Therefore, we have:

\[ \sqrt[n]{|a_n|} = \begin{cases}1 & \text{if } n = k^2, k \in \mathbb{N}_0 \\ 0 & \text{otherwise}\end{cases} \]

The sequence \(\sqrt[n]{|a_n|}\) is divergent, but its limit superior is \(1\):

\[ \limsup_{n\to \infty} \sqrt[n]{|a_n|} = 1 \]

Therefore, the radius of convergence is \(1\):

\[ r = \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}} = \frac{1}{1} = 1 \]

Theorem: Radius of Convergence via \(n\)-th Root Limit

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (x - c)^n\) be a real power series.

If the limit of \(\frac{1}{\sqrt[n]{|a_n|}}\) exists or is \(\infty\), then it is equal to the radius of convergence \(r\):

\[ r = \lim_{n \to \infty}\frac{1}{\sqrt[n]{|a_n|}} \in \mathbb{R}_{\ge 0} \cup \{\infty\} \]

Proof

TODO