2.4 Power Series
A Power Series: A series of the form \(\,a_0 + a_1z + a_2z^2 + \cdots \,\) where \(a_i\in \mathbb {C}\) is called a power series around 0 in the variable
\(z\).
- \(\implies \)
- Power series around \(0\) converges for \(z = 0\)
- \(\implies \)
- Power series around \(a\) converges for \(z = a\)
A power series is said to be convergent at a point \(z = z_0\) if the partial sums \(S_n\) evaluated at \(z_0\) converges to a
limit.
Example 2.17 (Geometric series). Determine for which \(z\) the series \(\displaystyle {\sum ^{\infty }_{n=0}z^n=1+z+z^2+\cdots }\) converges, and find its sum.
Solution. \(\displaystyle {1 + z + z^2 + \cdots }\)
\(\displaystyle {(1 - z)\big (1 + z + z^2 + \cdots + z^n\big ) = 1 - z^{n+1}}\)
\(\displaystyle {1 + z + z^2 + \cdots + z^n = \frac {1 - z^{n + 1}}{1 - z}}\,\) if \(z\neq 1\)
Since \(\lim \limits _{n\rightarrow \infty } z^{n + 1} = 0\) if \( \left |z\right | < 1\).
when \(\left |z\right | < 1\) the geometric series \(\displaystyle {\sum ^{\infty }_{n = 0}z^n}\) converges to \(\frac {1}{1 - z}\)
\(z^{n + 1}\) diverges for \(\left |z\right |> 1\).
So, \(\displaystyle {\sum z^n = \begin {cases} \text {converges}\, \frac {1}{1 - z}\,\text {when}\, \left |z\right | < 1\\\\ \text {diverges when}\,\left |z\right |> 1\\ \end {cases} }\)
Theorem 2.18. To each power series \(\displaystyle {\sum a_n z^n}\) there exists a corresponding \(R\) with \(0\leq R\leq \infty \) called the radius of convergence, with the following properties
- i).
- \(\displaystyle {\sum a_n z^n}\) converges absolutely for every \(z\) with \(\left |z\right | < R\).
- ii).
- If \(\left |z\right |> R\), the terms of the series are unbounded and hence the series is divergent.
Proof.
- i).
- Let \(R = \sup \big \{\left |z\right |:\, \sum \left |a_n z^n\right |\, \text {converges}\big \}\). If \(\left |z\right | < R\) then there is a \(z_1\ni \left |z\right | < \left |z_1\right | < R\) and \(\displaystyle {\left |a_nz_1^n\right |}\) converges.
Now there is an \(\,M\geq 0 \ni \left |a_nz_1^n\right |\leq M \, \forall n \geq 0\, \cdots \,(*) \quad \) then \begin {align*} \left |a_nz^n\right | & = \left |a_n\right |\,\left |\frac {z}{z_1}\right |^n\,\left |z_1\right |^n\\\\ & \leq M\,\left |\frac {z}{z_1}\right |^n\quad \text {by}\,(*) \end {align*}Since \(\,\displaystyle {\left |\frac {z}{z_1}\right | < 1}\)
\(\displaystyle {\sum M \left |\frac {z}{z_1}\right |^n = M \sum \left |\frac {z}{z_1}\right |^n}\,\) converges and so by comparison test \(\displaystyle {\sum a_n z^n}\) converges absolutely.
- ii).
- If \(\left |z\right |>R\) and if \(\displaystyle {a_nz^n}\) is convergent then there is an \(M\geq 0 \ni \left |a_n z^n\right | \leq M,\, \forall \, n\geq 0\).
So, \(\forall \, w \in \mathbb {C}\) with \(\left |z\right | > \left |w\right | >R\) \[\left |a_nw^n\right | \leq \left |a_nz^n\right |\,\left |\frac {w}{z}\right |^n \leq M\,\left |\frac {w}{z}\right |^n\] Since \(\displaystyle {\sum M\,\left |\frac {w}{z}\right |^n}\) is a convergent geometric series. \(\displaystyle {\sum \left |a_nw^n\right |}\) is also convergent by comparison test.
This is a contradiction to the definition of \(R\). So \(\displaystyle {\sum a_n z^n}\) is divergent for \(\left |z\right |> R\).
\(\star \) Determination of the radius of convergence of a given power series can be done using the ratio or the
root test.
Example 2.19. Consider the series \(\,\displaystyle {\sum ^{\infty }_{n = 0} \frac {z^n}{n!}}\)
\(\therefore \) By ratio test, this series converges absolutely when \begin {align*} \lim \limits _{n\rightarrow \infty } \left |\frac {z^{n+1}\big /\big (n+1\big )!}{z^n\big /n!}\right |& < 1\\\\ \lim \limits _{n \rightarrow \infty } \left |\frac {z}{n+1}\right | & < 1 \end {align*}
\(0 < 1 \,\) is always true independent for any \(z\in \mathbb {C}\), this converges absolutely. i.e the radius of convergence is
\(\infty \).
\(\star \) Cauchy\(-\)Hadamad formula for radius of convergence.
\(\therefore \) The series \(\displaystyle {\sum a_n z^n}\) has radius of convergence \(R\) where \(R\) is given by
\[\frac {1}{R} = \lim \sup \Big \{\sqrt [n]{\left |a_n\right |}\Big \}\]
\[\frac {1}{0} = \infty \,, \, \frac {1}{\infty }= 0\]
Example 2.21. Determine the radius of convergence \(R\) of the power series \(\, \displaystyle {\sum ^{\infty }_{n = 0} \frac {\big (2n\big )!}{\big (n!\big )^2}\,\big (z - 3i\big )^n}\)
\begin {align*} L^* & = \lim \limits _{n\rightarrow \infty } \left |\frac {\big (2n + 2\big )!}{\big ((n+1)!\big )^2}\div \frac {\big (2n)!}{\big (n!\big )^2}\right |\\\\ & = \lim \limits _{n\rightarrow \infty } \frac {\big (2n + 2\big )\big (2n+1\big )}{\big (n + 1\big )^2}\\\\ & = 4 \end {align*}
Hence \(R = \frac {1}{L^*} = \frac {1}{4}\). The series converges in the disc \(\left |z - 3i\right |< \frac {1}{4}\) of radius \(\frac {1}{4}\) and centre \(3i\).
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