ANA1_L31

Prologue

Definition 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).

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More convergence tests

Root test

Rudin, 3.33

Theorem 2(Theorem).

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 (in fact, it 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 $\sqrt[n]{|a_n|}<\beta$.

$$\begin{array}{rl}|a_n|& <{\beta }^n\\ |a_{n+1}|& <{\beta }^{n+1}\\ & ⋮\end{array}$$

So, for $n>N$, $|a_n|<{\beta }^n$. 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. ❏

Ratio test

Rudin, 3.34

Theorem 3(Theorem).

The series $\sum a_n$

1. converges if $\mathrm{limsup}|\frac{a_{n+1}}{a_n}|<1$,
2. 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.

Read examples at 3.35.

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Power series

Definition 4(Definition (Rudin, 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 5(Theorem (Rudin, 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 if $|z|<R$ and diverges if $|z|>R$.

Proof
Use the root test:

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

$R$ is called the convergence radius of $\sum c_n{z}^n$.

Example 6(Examples).

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 $p>1$).