# Week 2 - Series

## Definition

• In general, if we try to add the terms of an infinite sequence $\{a_n\}_{n=1}^{\infty}$ we get an expression of the form $$a_1+a_2+a_3+\cdots+a_n+\cdots$$
• which is called an infinite series (or just a series) and is denoted, for short, by the symbol $$\sum_{n=1}^{\infty} a_n \ \text{or}\ \sum{a_n}$$

### Convergent and Divergent

• Given a series $\sum_{n=1}^{\infty}a_n = a_1 + a_2 + a_3 + \cdots$, let $s_n$ denote its nth partial sum: $$s_n = \sum_{i=0}^n a_i = a_1 + a_2 + a_3 + \cdots + a_n$$
• If the sequence $\{s_n\}$ is convergent and $\lim_{n \to \infty} s_n = s$ exists as a real number, then the series $\sum a_n$ is called convergent and we write $$a_1 + a_2 + a_3 + \cdots + a_n + \cdots = s \ \text{or}\ \sum_{n=1}^{\infty} a_n = s$$
• The number s is called the sum of the series. If the sequence $\{s_n\}$ is divergent, then the series is called divergent.

## Geometric Series

• $a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} + \cdots = \sum_{n=1}^{\infty} ar^{n-1}, \ a \ne 0$
• Each term is obtained from the preceding one by multiplying it by the common ratio r.
• If $r = 1$, then $s_n = a + a + a + \cdots + a = na \to \pm \infty$. Since $\lim_{n \to \infty} s_n$ doesn’t exist, the geometric series diverges in this case.
• If $r \ne 1$, we have \begin{aligned} s_n &= a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} \ rs_n &= ar + ar^2 + ar^3 + \cdots + ar^{n} \end{aligned}
• Subtracting these equations, we get $$s_n - rs_n = a - ar^n$$
• $s_n = \frac{a(1-r^n)}{1-r}$

• If $-1 \le r \le 1$, then $r^n \to 0,\ \text{as}\ n \to \infty$, so $$\lim_{n \to \infty}s_n = \lim_{n \to \infty}\frac{a(1-r^n)}{1-r} = \frac{a}{1-r} - \frac{a}{1-r}\lim_{n \to \infty}r^n = \frac{a}{1-r}$$
• Summarize the results The geometric series $$\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \cdots$$ is convergent if $|r| < 1$ and its sum is $$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}\ \ \ \ |r| < 1$$ If $|r| \ge 1$, the geometric series is divergent.
• Another way to get the conclusion ($-1 \le r \le 1$):
• This figure provides a geometric demonstration of the result in the example. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles, $$\frac{s}{a} = \frac{a}{a- ar}\ \text{, so}\ s= \frac{a}{1-r}$$
• Example: $0.\bar{9}$
• \begin{aligned} 0.\bar{9} &= \sum_{n=1}^{\infty}9 \cdot 10^{-n} \\ &= 9 \sum_{n=1}^{\infty}10^{-n} = 9 \sum_{n=1}^{\infty}\frac{1}{10^n} \\ &= 9 \cdot \frac{1/10}{1- 1/10} = 9 \cdot \frac{1}{9} = 1 \end{aligned}

• Notice the formula is $ar^{n-1}$, so here $= 9 \cdot \frac{1/10}{1- 1/10}$

## Telescoping Series

• A telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation.
• For example, the series $$\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}$$
• (the series of reciprocals of pronic numbers) simplifies as {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}&{}=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\{}&{}=\lim _{N\to \infty }\sum _{n=1}^{N}\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\{}&{}=\lim _{N\to \infty }\left\lbrack {\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)}\right\rbrack \{}&{}=\lim _{N\to \infty }\left\lbrack {1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}}\right\rbrack \{}&{}=\lim _{N\to \infty }\left\lbrack {1-{\frac {1}{N+1}}}\right\rbrack =1.\end{aligned}}
• From Wikipedia

• \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$$is divergent. • Similarly, $s_{32} > 1 + \frac{5}{2}, s_{64} > 1 + \frac{6}{2}$, and in general$$s_{2^n} > 1 + \frac{n}{2}$$## Theorem • If the series $\displaystyle\sum_{n=1}^{\infty} a_n$ is convergent, then $\displaystyle\lim_{n \to \infty} a_n = 0$. • If $\displaystyle\lim_{n \to \infty} a_n$ does not exist or if $\displaystyle\lim_{n \to \infty} a_n \ne 0$, then the series $\displaystyle\sum_{n=1}^{\infty}a_n$ is divergent. • If $\sum a_n$ and $\sum b_n$ are convergent series, then so are the series $\sum c a_n$ (where c is a constant), $\sum(a_n + b_n)$, and $\sum(a_n - b_n)$, and • $\displaystyle\sum_{n=1}^{\infty} c a_n = c \sum_{n=1}^{\infty} a_n$ • $\displaystyle\sum_{n=1}^{\infty} (a_n + b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n$ • $\displaystyle\sum_{n=1}^{\infty} (a_n - b_n) = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n$ ## The Comparison Tests • The Comparison Test Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. • If $\sum b_n$ is convergent and $a_n < b_n$ for all n, then $\sum a_n$ is also convergent. • If $\sum b_n$ is divergent and $a_n > b_n$ for all n, then $\sum a_n$ is also divergent. • Example: Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges? • Let’s start with n = 2, which should get the same conclusion. • $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} < \sum_{n=1}^{\infty} \frac{1}{n \cdot (n-1)} = \sum_{n=1}^{\infty} (\frac{1}{n -1} - \frac{1}{n}) = \lim_{n \to \infty} 1 - \frac{1}{n} = 1$ • $0 < \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} < 1$ ## Cauchy Condensation • The sequence $\{a_k\}$ decreasing and $a_k > 0$. The series $\displaystyle\sum_{k=1}^{\infty} a_k$ converges if and only if $\displaystyle\sum_{k=0}^{\infty} 2^k \cdot a_{2^k}$ converges. • Proven: • $a_k > 0$, so $\displaystyle\sum_{k=1}^{\infty} a_k$ is non-decreasing. • Same as the Harmonic Series part, we can get:$$\begin{aligned}
s_n &= 1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7 + a_8) + \cdots \
s_n &< 1 + 2 \cdot a_2 + 4 \cdot a_4 + \cdots \
s_n &< \sum_{k=0}^{\infty} 2^k \cdot a_{2^k}
\end{aligned}
• Same example: Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges?
• First way -> Use Telescoping Series:
• $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge, iff $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ converge.
• $0 \le \frac{1}{n^2} \le \frac{1}{n^2 - n}\ \text{, }\ n \ge 2$ => if $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n}$, then $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ converge.
• $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n} = \sum_{n=2}^{\infty} (\frac{1}{n-1} - \frac{1}{n})$ which is a telescoping series.
• $\displaystyle \lim_{n \to \infty}\sum_{n=2}^{N} (\frac{1}{n-1} - \frac{1}{n}) = \lim_{n \to \infty}(1 - \frac{1}{N}) = 1$
• So $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n}$ converge => $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ converge => $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge
• Second way -> Use Conchy Condensation:
• The sequence $\{\frac{1}{n^2}\}$ decreasing and $\frac{1}{n^2} > 0$, So:
• $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} < \sum_{n=0}^{\infty} 2^n \cdot \frac{1}{(2^n)^2} = \sum_{n=0}^{\infty} \frac{1}{2^n} = 2$
• With Geometric Series $\displaystyle\sum_{n=0}^{\infty} \frac{1}{2^n} = \frac{1}{1-1/2} = 2$