LEC ANA1 31

Absolutely and conditionally convergent series

Definition 153.1(Definition).

We say $\sum a_k$ converges absolutely if $\sum |a_k|$ converges.

For example, $\sum a_k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ converges (by 3.43), but not absolutely. For a first principle proof, consider even and odd groupings of 2 consecutive terms, and see that they individually converge (recall the comparison test):

$$\begin{array}{rl}{s}_{2n}=\sum _{k=1}^{2n}{a}_k& =\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{2n-1}-\frac{1}{2n}\right)\\ & \le \frac{1}{1^2}+\frac{1}{3^2}+\cdots +\frac{1}{(2n-1{)}^2}\leftarrow \text{this converges}\\ & \\ s_{2n+1}=\sum _{k=1}^{2n+1}{a}_k& =1-\left(\frac{1}{2}-\frac{1}{3}\right)-\cdots -\left(\frac{1}{2n}-\frac{1}{2n+1}\right)\\ & \le 1-\left(\frac{1}{3^2}+\frac{1}{5^2}+\cdots +\frac{1}{(2n+1{)}^2}\right)\leftarrow \text{this converges}\end{array}$$

Let $(s_{2n})\to l_1$ and $(s_{2n+1})\to l_2$. Now, triangle inequality to the rescue:

$$\begin{array}{rl}|l_1-l_2|& =|l_1-s_{2n}|+|s_{2n}-s_{2n+1}|+|s_{2n+1}-l_2|\\ & =|l_1-s_{2n}|+\frac{1}{2n+1}+|s_{2n+1}-l_2|\end{array}$$

The RHS can be made less than $ϵ$ for any $ϵ$. Thus, $l_1=l_2\equiv l$. So, the sequence $s_1,s_2,s_3,\dots$ must also converge to $l$ (which is $\ln 2$, btw).

Important

If a convergent series is not absolutely convergent (in which case it is said to be conditionally convergent), we are not at liberty to rearrange the terms. This severely restricts the manipulations we can perform. On the other hand, the limit of an absolutely convergent series is independent of the order in which we sum the terms.

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More convergence tests

Theorem 153.2(Root test, Rudin (1976) 3.33).

Given $\sum a_n$, define

$$\alpha \equiv \underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|a_n|}.$$

Then,

1. if $\alpha <1$, $\sum a_n$ converges absolutely;

2. if $\alpha >1$, $\sum a_n$ diverges;

3. if $\alpha =1$, the test is inconclusive.

Proof.

If$\alpha <1$, pick $\alpha <\beta <1$. We know that there exists $N$ such that $n\ge N$ implies $\sqrt[n]{|a_n|}<\beta$, i.e, $|a_n|<{\beta }^n$. We know $\sum {\beta }^n$ converges. Hence, from the comparison test, $\sum |a_n|$ converges. This implies $\sum a_n$ converges.

If $\alpha >1$, we know that for all $N$, there exists $n>N$ such that $\sqrt[n]{|a_n|}>1$, i.e, $|a_n|>1$. So, ${\lim }_{n\to \infty }|a_n|\ne 0$.

To prove (3), consider series $\sum \frac{1}{n}$ and $\sum \frac{1}{n^2}$. Both have $\alpha =1$ (Rudin, 3.20), but the first diverges, while the second converges.□

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Theorem 153.3(Ratio test, Rudin (1976) 3.34).

The series $\sum a_n$
4. converges if $\mathrm{limsup}|\frac{a_{n+1}}{a_n}|<1$,
5. diverges if $|\frac{a_{n+1}}{a_n}|\ge 1$ for all $n\ge N_0$, where $N_0$ is some fixed integer.

Proof.

If (1) holds, in the same manner as in the previous proof, we can pick$\mathrm{limsup}|\frac{a_{n+1}}{a_n}|<\beta <1$ such that there exists $N$ such that for all $n>N$, $|\frac{a_{n+1}}{a_n}|<\beta$. In particular,

$$\begin{array}{rl}|a_{N+1}|& <\beta |a_N|\\ |a_{N+2}|& <\beta |a_{N+1}|<{\beta }^2|a_N|\\ & ⋮\\ |a_{N+p}|& <{\beta }^p|a_N|\end{array}$$

So, $|a_n|<(|a_N|{\beta }^{-N}){\beta }^n$ for all $n\ge N$. The result follows from the comparison test.

(2) prevents the limit of individual terms of the sequence from being 0.□

Warning

Note that $\mathrm{limsup}|\frac{a_{n+1}}{a_n}|\ge 1$ for all $n\ge N_0$ for some fixed $N_0$ does not guarantee divergence. A counter example can be easily constructed by interlacing two appropriate convergent series.

See Rudin (1976) 3.35 for examples.

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Power series

Definition 153.4(Rudin (1976) 3.38).

The power series of a sequence $(c_n)$ of complex numbers is the series

$$\sum _{n=0}^{\infty }{c}_n{z}^n.$$

Theorem 153.5(Rudin (1976) 3.39).

Given the power series $\sum c_n{z}^n$, define

$$\alpha \equiv \underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|c_n|},R\equiv \frac{1}{\alpha }$$

If $\alpha =0$, $R=\infty$ and if $\alpha =\infty$, $R=0$. Then, $\sum c_n{z}^n$ converges absolutely if $|z|<R$ and diverges if $|z|>R$. $R$ is called the convergence radius of $\sum c_n{z}^n$.

Proof.

Use the root test:

$$\begin{array}{r}\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|c_n{z}^n|}=|z|\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|c_n|}=\frac{|z|}{R}\end{array}$$ □

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Remark 153.6(Remark).

In Thm 5, the convergence of $\sum c_n{z}^n$ is uniform on every closed disc of radius $r<R$ centered at the origin, as you can show using Thm 155.12. Thus, it follows that the series converges uniformly on every compact subset of the convergence disc. Uniform convergence on the whole disc fails in general.

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Example 153.7(Example).

1. The series $\sum n^n{z}^n$ has $R=0$, i.e, it converges only when $z=0$.
2. The series $\sum \frac{z^n}{n^n}$ has $R=\infty$, i.e, it converges for every $z$.
3. The above theorem is not easily applicable on the series $\sum \frac{z^n}{n!}$, since showing $\lim \sup \sqrt[n]{1/n!}=0$ takes some work. It is much easier to use the ratio test: $\lim \sup |z|/(n+1)=0$, for all $z$. Thus, $R=\infty$.
4. The series $\sum z^n$ has $R=1$. If $|z|=1$, the series diverges, since $z^n$ does not tend to $0$ as $n\to \infty$.
5. Thanks to the limits we proved here, it follows that $\sum z^n/n$ and $\sum z^n/n^2$ have $R=1$. The former converges for all $z$ with magnitude 1 except $z=1$ (TBP). The latter converges for all $z$ with magnitude 1 by the comparison test, since $|z^n/n^2|=1/n^2$ ($1/n^2$ converges since it is a p-series with $p>1$).

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References

Rudin, W. (1976). Principles of Mathematical Analysis (3d ed). McGraw-Hill.