Home page for accesible maths 2 Series 2.1 Series versus sequence 2.3 Convergence tests

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2.2 Some important examples of series

Proposition 2.2.1 (The geometric series).

If $|x|<1$ then $\sum_{n=0}^{\infty}x^{n}$ converges and

\sum_{n=0}^{\infty}x^{n}=1+x+x^{2}+\cdots=\frac{1}{1-x}.

Proof.

Let $A_{n}=1+x+\ldots+x^{n}$ . Then $xA_{n}=x+x^{2}+\ldots+x^{n+1}$ , so $A_{n}-xA_{n}=1-x^{n+1}$ (the other terms cancel). Now $x^{n}\to 0$ as $n\to\infty$ because $|x|<1$ , so we find

A_{n}=\frac{1-x^{n+1}}{1-x}\rightarrow\frac{1}{1-x},\quad n\rightarrow\infty.

∎

Key idea of proof.

study $A_{n}-xA_{n}$

Let us look at a few applications of this.

Example 2.2.2.

(1)

$\displaystyle\sum_{n=0}^{\infty}\frac{2^{n}}{3^{n}}=\sum_{n=0}^{\infty}\Big{(}% \frac{2}{3}\Big{)}^{n}=\frac{1}{1-\frac{2}{3}}=3.$
(2)

$\displaystyle\sum_{n=0}^{\infty}x^{2n+1}=x(1+x^{2}+x^{4}+\cdots)=\frac{x}{1-x^% {2}}$ whenever $|x|<1$ .
(3)

$\displaystyle\frac{1}{1+y^{2}}=1-y^{2}+y^{4}-y^{6}+\cdots$ whenever $|y|<1$ (replacing $x$ by $-y^{2}$ ).
(4)

If $|x|<|a|$ then

$\frac{1}{a-x}=\frac{1}{a}\,\frac{1}{1-x/a}=\frac{1}{a}\Big{(}1+\frac{x}{a}+% \frac{x^{2}}{a^{2}}+\cdots\Big{)}=\frac{1}{a}+\frac{x}{a^{2}}+\frac{x^{2}}{a^{% 3}}+\cdots.$
(5)

The expression $0{\cdot}\dot{2}\dot{7}$ means $0{\cdot}272727\cdots$ . It equals

$\frac{27}{100}\Big{(}1+\frac{1}{100}+\frac{1}{10000}+\cdots\Big{)}=\frac{27}{1% 00}\,\frac{1}{1-\frac{1}{100}}=\frac{27}{99}=\frac{3}{11}.$

Proposition 2.2.3.

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ converges to $1$ .

Proof.

Let $\displaystyle a_{n}=\frac{1}{n(n+1)}$ . Then $\displaystyle a_{n}=\frac{1}{n}-\frac{1}{n+1}$ , so for the $n$ -th partial sum we find

A_{n}=a_{1}+a_{2}+\ldots+a_{n}=\Big{(}1-\frac{1}{2}\Big{)}+\Big{(}\frac{1}{2}-% \frac{1}{3}\Big{)}+\cdots+\Big{(}\frac{1}{n}-\frac{1}{n+1}\Big{)}=1-\frac{1}{n% +1}.

Thus

A_{n}=1-\frac{1}{n+1}\rightarrow 1,\quad n\rightarrow\infty,

proving our claim. ∎

The second important example, the “harmonic series”, shows that the converse of Proposition 2.1.4(1) is false: one can have $a_{n}\rightarrow 0$ but the series $\sum_{n=1}^{\infty}a_{n}$ being divergent.

Proposition 2.2.4 (The harmonic series).

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent.

Proof.

Combine the terms of sums into brackets, as follows:

1+\frac{1}{2}+\Big{(}\frac{1}{3}+\frac{1}{4}\Big{)}+\Big{(}\frac{1}{5}+\ldots+% \frac{1}{8}\Big{)}+\ldots.

So we can write the $2^{n}$ -th partial sum as

A_{2^{n}}=1+\frac{1}{2}+\Big{(}\frac{1}{3}+\frac{1}{4}\Big{)}+\ldots\Big{(}% \frac{1}{2^{n-1}+1}+\ldots+\frac{1}{2^{n}}\Big{)}

Each bracket has sum at least $\frac{1}{2}$ and there are $n+1$ such brackets, so

A_{2^{n}}\geq\frac{n+1}{2}.

Now $(A_{n})_{n\in\mathbb{N}}$ is clearly increasing, and

\frac{n+1}{2}\rightarrow\infty,\quad n\rightarrow\infty,

which shows that $(A_{2^{n}})_{n\in\mathbb{N}}$ and hence the sequence of partial sums $(A_{n})_{n\in\mathbb{N}}$ tends to $\infty$ , so the series diverges. ∎

Key idea of proof.

combine summands into suitable groups

Comment: If a sequence $(a_{n})_{n\in\mathbb{N}}$ converges to $0$ then the corresponding series $\sum_{n=1}^{\infty}a_{n}$ may converge or diverge. If a sequence $(a_{n})_{n\in\mathbb{N}}$ does not converge to $0$ then the series $\sum_{n=1}^{\infty}a_{n}$ must diverge.

Exercise X.

Find three non-convergent series that Proposition 2.1.5 does not detect.

Exercise W2.3.

Suppose a series $\sum_{n=1}^{\infty}a_{n}$ converges and another series $\sum_{n=1}^{\infty}b_{n}$ diverges. What can you say about the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ ? If you can make general statements then prove them, otherwise provide examples or counter-examples.