Mathematical Analysis
Lecture 5
3.1 Series
Definition 3.1.1. Given a sequence $\{ x_n \}$, we write the formal object
$\ds \sum_{n=1}^\infty x_n $ or just $\ds\sum x_n $ and call it a series.
A series converges if the sequence $\{ s_k \}$ defined by $s_k := \displaystyle\sum_{n=1}^k x_n = x_1 + x_2 + \cdots + x_k ,$ converges. The numbers $s_k$ are called partial sums. If $x := \lim\, s_k$, we write \begin{equation*} \sum_{n=1}^\infty x_n = x . \end{equation*}
3.1 Series
Sequence of Partial sums
Example 3.1.1.
Claim: The series $\ds\sum_{n=1}^\infty \frac{1}{2^n}$ converges and
the limit is 1.
That is, $\ds\sum_{n=1}^\infty \frac{1}{2^n} $
$\ds =
\lim_{k\to\infty} \sum_{n=1}^k \frac{1}{2^n} $
$ = 1 .$
Proof.
First we prove the following equality \begin{equation*}
\sum_{n=1}^k \frac{1}{2^n} + \frac{1}{2^k} =
1 . \end{equation*}
Example 3.1.1.
👉 $\;\ds \sum_{n=1}^k \frac{1}{2^n}+ \frac{1}{2^k} = 1 .$
The equality is true when $k=1$. Assume it is true for $k.$ We need to prove for $k+1.$ Thus we start with
$ \ds \sum_{n=1}^{k+1} \frac{1}{2^n} + \frac{1}{2^{k+1}} $ $ \ds = \left( \sum_{n=1}^{k} \frac{1}{2^n} + \frac{1}{2^{k+1}} \right) + \frac{1}{2^{k+1}} \qquad \qquad$
$ \qquad\ds = \sum_{n=1}^{k} \frac{1}{2^n} + \frac{2}{2^{k+1}} $ $ \ds = \sum_{n=1}^{k} \frac{1}{2^n} + \frac{1}{2^{k}} $ $ \ds = 1 . $
😃
Example 3.1.1.
👉 $\;\ds \sum_{n=1}^k \frac{1}{2^n} + \frac{1}{2^k} = 1 .$
We write $\ds \sum_{n=1}^k \frac{1}{2^n} $ $\ds= 1 - \frac{1}{2^k} .$ Then
$\ds \lim_{k\to \infty } \sum_{n=1}^k \frac{1}{2^n} $ $\ds= \lim_{k\to \infty }1 - \lim_{k\to \infty }\frac{1}{2^k} $ $\ds= 1 . $
Therefore, the series $\ds\sum_{n=1}^\infty \frac{1}{2^n}$ converges and its limit is 1.
3.1 Series
Theorem 3.1.1. If $x_n \geq 0$ for all $n$, then $\sum x_n$ converges if and only if the sequence of partial sums is bounded above.
3.1 Series
Theorem 3.1.1. If $x_n \geq 0$ for all $n$, then $\sum x_n$ converges if and only if the sequence of partial sums is bounded above.
Proof. $\nec$ By definition, $\sum x_n$ converges as a series if and only if $\{s_k \}$, with $s_k = \sum_{n=1}^{k} x_n$, converges as a sequence. If $\{s_k \}$ converges, then it is bounded, by Theorem 2.1.2.
$\suf$ Now suppose that the sequence of partial sums is bounded above. Notice that
$ s_{k+1} $ $ = x_1 + x_2 + \cdots + x_k+ x_{k+1} $ $ = s_k + x_{k+1} $ $ \geq s_k $
since $x_n \geq 0$.
3.1 Series
Theorem 3.1.1. If $x_n \geq 0$ for all $n$, then $\sum x_n$ converges if and only if the sequence of partial sums is bounded above.
Proof. $\suf$ So we have \[ s_{k+1} \geq s_k \,\text{ for all }\, k\in \N. \]
Then $\{s_k \}$ is monotone increasing. If $\{s_k \}$ is bounded above, then by the Monotone Convergence Theorem (Theorem 2.2.1), it converges. $\;\bs$
3.1.1 Cauchy series
Definition 3.1.2 A series $\sum x_n$ is said to be Cauchy or a Cauchy series if the sequence of partial sums $\{ s_n \}$ is a Cauchy sequence.
3.1.1 Cauchy series
Theorem 3.1.2 The series $\sum x_n$ is Cauchy if for every $\vre > 0$, there exists an $M \in \N$ such that for every $n \geq M$ and every $k \gt n$, we have \begin{equation*} \abs{ \sum_{j={n+1}}^k x_j } \lt\vre . \end{equation*}
Example 3.1.2.
Claim: The series $\ds\sum \frac{1}{n}$ diverges. a.k.a. harmonic series.
Example 3.1.2.
Claim: The series $\ds\sum \frac{1}{n}$ diverges. a.k.a. harmonic series.
Proof. Consider the partial sums $s_n$ for $n=2^k$:
$s_1 = 1,\;\;$ $\ds s_2 = \left( 1 \right) + \left( \frac{1}{2} \right) ,$
$\ds s_4 = \left( 1 \right) + \left( \frac{1}{2} \right) + \left( \frac{1}{3} + \frac{1}{4} \right) ,$
$\ds s_8 = \left( 1 \right) + \left( \frac{1}{2} \right) + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) ,$
Example 3.1.2.
Claim: The series $\ds\sum \frac{1}{n}$ diverges. a.k.a. harmonic series.
Proof. Consider the partial sums $s_n$ for $n=2^k$:
$\ds s_{2^k} = 1 + \sum_{j=1}^k \left( \sum_{m=2^{j-1}+1}^{2^j} \frac{1}{m} \right) . $
Note that $\;\ds\frac{1}{3} + \frac{1}{4}$ $\ds \geq \frac{1}{4} + \frac{1}{4} $ $\ds= \frac{1}{2} $
and $\;\ds \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} $ $\ds\geq \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$ $\ds= \frac{1}{2}. $
Example 3.1.2.
$\ds s_{2^k} = 1 + \sum_{j=1}^k \left( \sum_{m=2^{j-1}+1}^{2^j} \frac{1}{m} \right) . $
Proof. In general
$\ds \sum_{m=2^{k-1}+1}^{2^k} \frac{1}{m} $ $\ds \geq \sum_{m=2^{k-1}+1}^{2^k} \frac{1}{2^k} $ $=\ds\left(2^{k-1}\right) \frac{1}{2^k} $ $\ds= \frac{1}{2} .$
$\Rightarrow \,\ds s_{2^k} $ $\ds \geq 1 + \sum_{j=1}^k \frac{1}{2} $ $\ds = 1 + \frac{k}{2} .$
Since $\left\{ \frac{k}{2} \right\}$ is unbounded, $\left\{ s_{2^k} \right\}$ is unbounded, and therefore $\{ s_n \}$ is unbounded.
Example 3.1.2.
Claim: The series $\ds\sum \frac{1}{n}$ diverges. a.k.a. harmonic series.
Proof.
Since $\left\{ \frac{k}{2} \right\}$ is unbounded, $\left\{ s_{2^k} \right\}$ is unbounded, and therefore $\{ s_n \}$ is unbounded.
Hence $\{ s_n \}$ diverges, and consequently $\ds\sum \dfrac{1}{n}$ diverges. $\,\blacksquare$
3.1.2 Basic properties
Theorem 3.1.3. Let $\sum x_n$ be a convergent series. Then the sequence $\{ x_n \}$ is convergent and \begin{equation*} \lim_{n\to\infty} x_n = 0. \end{equation*}
3.1.2 Basic properties
Theorem 3.1.3. Let $\sum x_n$ be a convergent series. Then the sequence $\{ x_n \}$ is convergent and $ \lim_{n\to\infty} x_n = 0.$
Proof. Let $\vre > 0$ be given. As $\sum x_n$ is convergent, it is Cauchy. Thus we find an $N$ such that for every $n \geq N$, we have
$ \ds \vre > \abs{ \sum_{j={n+1}}^{n+1} x_j } $ $ = \abs{ x_{n+1} } .$
Therefore for every $n \geq N+1$, we have $\abs{x_{n}} \lt \vre.\; \blacksquare$
3.1.2 Basic properties
$\sum x_n$ convergent $\implies \lim x_n =0$
Example 3.1.3
If $r \geq 1$ or $r \leq -1$, then the geometric series
$\qquad\qquad \qquad \ds\sum_{n=0}^\infty r^n$ diverges.
Proof.
Notice that $\abs{r^n} = \abs{r}^n$
$ \geq 1^n$
$= 1.$
So the terms do not go to zero
and, by the previous theorem, the series cannot converge.
$\;\blacksquare$
3.1.2 Basic properties
Theorem 3.1.4. (Linearity of series) Let $\alpha \in \R$ and $\sum x_n$ and $\sum y_n$ be convergent series. Then
- $\sum \alpha x_n$ is a convergent series and \begin{equation*} \quad\qquad\sum_{n=1}^\infty \alpha x_n = \alpha \sum_{n=1}^\infty x_n . \end{equation*}
- $\sum ( x_n + y_n )$ is a convergent series and \begin{equation*} \qquad\qquad \sum_{n=1}^\infty ( x_n + y_n ) = \left( \sum_{n=1}^\infty x_n \right) + \left( \sum_{n=1}^\infty y_n \right) . \end{equation*}
👀 Complementary reading 📖
3.1.2 Basic properties
Geometric series
Theorem 3.1.5. Suppose $-1 \lt r \lt 1$. Then the geometric series $\sum_{n=0}^\infty r^n$ converges, and \begin{equation*} \sum_{n=0}^\infty r^n = \frac{1}{1-r} . \end{equation*}
👀 Complementary reading 📖
3.1.2 Basic properties
Geometric series $\ds\sum_{n=0}^\infty r^n = \frac{1}{1-r}$