Mathematics Advanced: Sequences and Series

• There are 3 types of sequences: arithmetic, geometric, and summing
• Sequences are also known as progressions

Arithmetic Sequences

• In arithmetic sequences, the difference between successive terms is constant:

  Examples:

  $d=T_n -T_{n-1},n\geq 2$

  $T_n -a+(n-1)d \text{ where } a=T_1$

• Suppose 3 numbers, $\text{a, b, }$and $\text{m}$, form an arithmetic sequence when $m-a=b-m$, which can be rearranged as $m=\frac{a+b}{2}$:
  • $m$ is called the arithmetic mean of $a$ and $b$

Geometric Sequences

• Geometric series have a non-zero ratio between successive terms:

Examples:

$\frac{T_n}{T_{n-1}}=r,n\geq 2$

$T_n =a\cdot r^{n-1} \text{ where }a=T_{1}$

• Three numbers, $\text{a, b, }$and $\text{m}$, form a geometric progression, when $\frac{b}{g}=\frac{g}{a}$, and therefore when $g^2 =a\cdot b$
• $g$ is the geometric mean of a, b

Summing Sequences

• Summing sequences occur when an infinite number of terms are added together: $$S_n =T_1 +T_2 +T_3 +…+ T_n$$
• Summing sequences can be represented using sigma notation:

$$\sum_{k=1}^{n} 2^{-n} = T_1 +T_2 +T_3 +…+ T_n =S_n$$

Sum of an Arithmetic Progression

Let $l=T_n$ be the last term of an AP with $T_1 =a$ and difference $d$. It can be therefore determined that:

• $S_n=a+(a+d)+(a+2d)+(a+3d)+…+(l-2d)+(l-d)+l$
• $S_n = l + (l−d) + (l − 2d ) + . . . + (a + 2d ) + (a + d ) + a$
• Combining the two equations gives $2S_n =n(a+l)$, as there are $n$ terms
• Therefore, $S_n =\frac{1}{2}n(a+l)$
• Since $l=T_n =a+(n-1)d,$ we can derive an alternative formula:
  • $S_n =\frac{1}{2}n(2a+(n-1)d)$
  • $\therefore S_n =\sum_{k=1}^{n}(a+(k-1)d)$

Sum of a Geometric Progession:

To find the sum of a geometric sequence:

• $S_{n}=a+ar+ar^{2} +…+ar^{n-2}+ar^{n-1}$
• $r\cdot S_{n}=ar+ar^{2}+ar^{3} +…+ar^{n-1}+ar^{n}$

Subtract the second equation from the first:

• $(r-1)S_n =ar^{n}-a$
• $S_n=\frac{a(r^{n}-1)}{r-1} \text{ when }\mid r\mid\lt 1$

For when $\mid r\mid\gt 1,$ subtract the first equation from the second:

• $(1-r)S_{n}=a-ar^{n}$
• $S_{n} =\frac{a(1-r^n)}{1-r} \text{ when }\mid r\mid\gt 1$

Limiting Sum

If $\mid r\mid\lt 1,$ then $\displaystyle{\lim_{n\to\infty}r^{n}=0},$ therefore:

• $\displaystyle{\lim_{n\to\infty}T_{n}=0}$
• $S_{\infty}=\displaystyle{\lim_{n\to\infty}S_{n}=\frac{a}{1-r}}$
• Progressions with limiting sums are said to converge on a value
  • For example, $\sum^{\infty}_{n=1}ar^{n-1} \text{ converges to }\frac{a}{1-r}$

Representing Recurring Decimals

If we want to express $1.1\overline{037}$ as a fraction, we can write it as a Geometric Progression $(1.1+0.0037+0.0000037+…):$

• In the brackets, there is a limiting sum: $S_{\infty}=\frac{0.0037}{1-0.001}$
• $s_{\infty}=\frac{1}{270}$
• BUT we need to add 1.1 to this sum:
• $\therefore 1.1\overline{037}=\frac{149}{135}$