Created on July 09, 2023

Last modified on July 10, 2023

Written by Some author

Read time: 3 minutes

Summary: A power series converges if the absolute value of the variable is less than the radius of convergence, determined by the limit supremum of the coefficients; the sum of a product series can be computed using partial sums and their differences; if the partial sums of a series are bounded and the coefficients are non-increasing and converge to zero, the product series converges; an alternating, absolutely decreasing, and converging-to-zero sequence leads to a convergent series; and if the radius of convergence of a power series is 1, the coefficients are non-increasing and converge to zero, the series converges for all points on the unit circle except possibly at 1.

$\textbf{Theorem}$ Given the power series $\Sigma c_n z^n$, put

$$ \alpha=\limsup _{n \rightarrow \infty} \sqrt[n]{\left|c_n\right|}, \quad R=\frac{1}{\alpha} . $$

(If $\alpha=0, R=+\infty$; if $\alpha=+\infty, R=0$.) Then $\Sigma c_n z^n$ converges if $|z|<R$, and diverges if $|z|>R$.

Proof:

Let $L= \limsup_{n \to \infty} \left(|c_nz^n| \right)^{1/n}= |z| \limsup_{n \to \infty} |c_n|^{1/n} = |z|\alpha $

If $L<1$, then $|z| < \alpha$, converges

If $L>1$, then $|z| > \alpha$, diverges

and we can't conclude anything when $L =1$.

$\textbf{Example}$ we can show that series $\sum \frac{z^n}{\Gamma(n+1)}$ has $R = +\infty$.

First note that $$\limsup_{n \to \infty } \left(|z^n/\Gamma(n+1)|\right)^{1/n} = |z|\limsup_{n \to \infty } \Gamma(n+1)^{-1/n}$$

$$= |z| \limsup_{n \to \infty }((2\pi n)^{1/2} (n/e)^n)^{-1/n} = |z| \lim_{n \to \infty }(2\pi n )^{-\frac{1}{2n}}(n/e)^{-1} = 0$$

$\textbf{Theorem}$ Given two sequences $\left\{a_n\right\},\left\{b_n\right\}$, put

$$ A_n=\sum_{k=0}^n a_k $$

if $n \geq 0$; put $A_{-1}=0$. Then, if $0 \leq p \leq q$, we have

$$ \sum_{n=p}^q a_n b_n=\sum_{n=p}^{q-1} A_n\left(b_n-b_{n+1}\right)+A_q b_q-A_{p-1} b_p $$.

Proof:

Consider the following:

$$\sum_{n=p}^q a_n b_n = \sum_{n=p}^q (A_n - A_{n-1}) b_n$$

$$= \sum_{n=p}^q A_n b_n - \sum_{n=p-1}^{q-1} A_n b_{n+1}$$

$$= \sum_{n=p}^{q-1} A_n (b_n - b_{n+1}) + A_{q}b_q -A_{p-1}b_{p}$$

$\textbf{Theorem}$ Suppose

(a) the partial sums $A_n$ of $\Sigma a_n$ form a bounded sequence;

(b) $b_0 \geq b_1 \geq b_2 \geq \cdots$;

Then $\Sigma a_n b_n$ converges.

Proof:

Since the partial sum $A_n$ of series $\sum a_n$ form. a bounded sequence, we will have $|A_n| < M$ for all $n$. Choose $N$ big enouch so that $|b_n| < \frac{\epsilon}{2M}$ for all $n > N$.

Then we will have the following for the partial sum $\sum_{i=p}^q a_nb_n$ for $p,q > n$.

$$\left|\sum_{i=p}^q a_nb_n\right| = \left|\sum_{i=p}^{q-1} A_n(b_{n}- b_{n+1}) +A_{q}b_{q}-A_{p-1}b_{p}\right| \le M \left|\sum_{i=p}^{q-1} (b_{n}- b_{n+1}) +b_{q}+b_{p}\right| =2 M \left|b_q \right|< \epsilon$$

So it follows from Cauchy criteria.

$\text{Comment}$. Alternating, absolutely decreasing and limiting to zero sequence has convergent series.

$\text{Theorem.}$ Suppose the radius of convergence of $\Sigma c_n z^n$ is 1 , and suppose $c_0 \geq c_1 \geq c_2 \geq \cdots, \lim _{n \rightarrow \infty} c_n=0$. Then $\Sigma c_n z^n$ converges at every point on the circle $|z|=1$, except possibly at $z=1$.

Proof. take $a_n = z^n$ and $b_n = c_n$ as of the above theorem. Then we only need partial $\sum z^n$ to be bounded.

Indeed, we have $|\sum_{i=0}^kz^n| = \left|\frac{1-z^{k+1}}{1-z}\right| \le \frac{1 + |z|^{k+1}}{\left| 1 -z \right|} = \frac{2}{\left| 1 -z \right|} $ since $|z| =1$ and unbounded when $z=1$, which we can't conclude anything based on above theorem