Series convergence

A series is studied through its sequence of partial sums

$$s_n=\sum _{k=0}^n{a}_k$$

so series convergence is a special case of sequence convergence.

For worked examples, see Convergence test examples.

For a series ${\sum }_{n=0}^{\infty }{a}_n$:

• it is absolutely convergent if ${\sum }_{n=0}^{\infty }|a_n|$ converges
• it is conditionally convergent if ${\sum }_{n=0}^{\infty }{a}_n$ converges, but ${\sum }_{n=0}^{\infty }|a_n|$ does not

Riemann Rearrangement Theorem

If a real series is conditionally convergent, then for every $L\in \mathbb{R}$ there is a bijection $\phi :{\mathbb{N}}_0\to {\mathbb{N}}_0$ such that

$$\sum _{n=0}^{\infty }{a}_{\phi (n)}=L$$

So a conditionally convergent series can be rearranged to converge to any prescribed real value. In contrast, absolutely convergent series keep the same sum under every rearrangement.

Leibniz Criterion

If $(a_n{)}_{n\in {\mathbb{N}}_0}$ is a monotone decreasing sequence of non-negative real numbers and $a_n\to 0$, then the alternating series

$$\sum _{n=0}^{\infty }(-1{)}^n{a}_n$$

converges.

Cauchy Criterion

The series ${\sum }_{n=0}^{\infty }{a}_n$ converges if and only if for every $\epsilon >0$ there exists $N\in {\mathbb{N}}_0$ such that for all $n\ge m\ge N$,

$$\left|\sum _{k=m+1}^n{a}_k\right|<\epsilon$$

So the tails of the series must become arbitrarily small.

p-Series Test

The series

$$\sum _{n=1}^{\infty }\frac{1}{n^p}$$

converges if and only if $p>1$.

Proof:

If $p>1$, split the series into dyadic blocks:

$$\sum _{n=1}^{\infty }\frac{1}{n^p}=1+\sum _{m=0}^{\infty }\sum _{n=2^m}^{2^{m+1}-1}\frac{1}{n^p}$$

For $2^m\le n<2^{m+1}$, we have

$$\frac{1}{n^p}\le \frac{1}{(2^m{)}^p}$$

and there are $2^m$ terms in this block. Hence

$$\sum _{n=2^m}^{2^{m+1}-1}\frac{1}{n^p}\le 2^m\cdot \frac{1}{(2^m{)}^p}=2^{-m(p-1)}$$

Therefore

$$\sum _{n=1}^{\infty }\frac{1}{n^p}\le 1+\sum _{m=0}^{\infty }{2}^{-m(p-1)}$$

and the right-hand side is a convergent geometric series because $p-1>0$.

If $p=1$, then for $2^m\le n<2^{m+1}$ we have

$$\frac{1}{n}\ge \frac{1}{2^{m+1}}$$

so

$$\sum _{n=2^m}^{2^{m+1}-1}\frac{1}{n}\ge 2^m\cdot \frac{1}{2^{m+1}}=\frac{1}{2}$$

Thus each dyadic block contributes at least $\frac{1}{2}$, so the harmonic series diverges.

If $0<p<1$, then $n^p\le n$ for all $n\ge 1$, hence

$$\frac{1}{n^p}\ge \frac{1}{n}$$

Since the harmonic series diverges, ${\sum }_{n=1}^{\infty }\frac{1}{n^p}$ also diverges by comparison.

If $p\le 0$, then

$$\frac{1}{n^p}=n^{-p}$$

does not converge to $0$, so the series cannot converge.

Limit Comparison Test

Let $a_n,b_n>0$ for all sufficiently large $n$. Assume

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=c$$

• If $0<c<\infty$, then $a_n$ and $b_n$ have the same size for large $n$. In this case, $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

• If $c=0$, then $a_n$ is asymptotically smaller than $b_n$ for large $n$. If $\sum b_n$ converges, then $\sum a_n$ also converges.

• If $c=\infty$, then $a_n$ is asymptotically larger than $b_n$ for large $n$. If $\sum b_n$ diverges, then $\sum a_n$ also diverges.

This is the same asymptotic idea used in Algorithm Runtime Analysis.

Root Test

Let

$$\rho =\underset{n\to \infty }{\mathrm{limsup}}|a_n{|}^{1/n}$$

Then:

• if $\rho <1$, the series ${\sum }_{n=0}^{\infty }{a}_n$ converges absolutely
• if $\rho >1$, the series diverges

Ratio Test

If $a_n\ne 0$ for all $n$ and

$$\rho =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$$

then:

• if $\rho <1$, the series ${\sum }_{n=0}^{\infty }{a}_n$ converges absolutely
• if $\rho >1$, the series diverges

Cauchy Product

If ${\sum }_{n=0}^{\infty }{a}_n$ and ${\sum }_{n=0}^{\infty }{b}_n$ are absolutely convergent, then

$$\left(\sum _{n=0}^{\infty }{a}_n\right)\left(\sum _{n=0}^{\infty }{b}_n\right)=\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{a}_{n-k}{b}_k\right)$$

and the series on the right also converges absolutely.

Power Series

A power series centered at $a$ has the form

$$\sum _{k=0}^{\infty }{c}_k(x-a{)}^k$$

It has a radius of convergence $R$ such that:

• it converges for $|x-a|<R$
• it diverges for $|x-a|>R$

The interval $(a-R,a+R)$ is called the interval of convergence. If

$$\rho =\underset{n\to \infty }{\mathrm{limsup}}|c_n{|}^{1/n}$$

then

$$R=\{\begin{array}{ll}0,& \rho =\infty \\ {\rho }^{-1},& 0<\rho <\infty \\ \infty ,& \rho =0\end{array}$$

when using root test on power series: $a_k=c_k(x-a{)}^k⟺\frac{a_k}{c_k}=(x-a{)}^k$

Example

Consider

$$\sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}{x}^n$$

This is a power series centered at $a=0$ with coefficients

$$c_n=(-1{)}^{n+1}\frac{1}{n}$$

Hence

$$|c_n{|}^{1/n}={\left(\frac{1}{n}\right)}^{1/n}\to 1$$

so

$$\rho =\underset{n\to \infty }{\mathrm{limsup}}|c_n{|}^{1/n}=1$$

and therefore

$$R={\rho }^{-1}=1$$

For the terms

$$a_n(x)=(-1{)}^{n+1}\frac{1}{n}{x}^n$$

the root test gives

$$\underset{n\to \infty }{\mathrm{limsup}}|a_n(x){|}^{1/n}=\underset{n\to \infty }{\mathrm{limsup}}{\left(\frac{1}{n}|x{|}^n\right)}^{1/n}=|x|$$

Therefore:

• if $|x|<1$, the series converges absolutely
• if $|x|>1$, the series diverges
• if $|x|=1$, the root test is inconclusive

Check the endpoints separately:

• at $x=1$, we get ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}$, which converges by the Leibniz Criterion
• at $x=-1$, we get $-{\sum }_{n=1}^{\infty }\frac{1}{n}$, which diverges

So the interval of convergence is

$$(-1,1]$$ \frac{1}{2^{n}} \text{n is even} \\ \frac{-1}{3^{n}} \text{n is odd} \end{cases}