Sequences & Series — Mathslive.ie

Section 1

Patterns to sequences

A plumber charges a €30 call-out fee and €20 per hour. We can build a table:

$x = 0$ hrs ⟹ €30

$x = 1$ hr ⟹ €50

$x = 2$ hrs ⟹ €70

$x = 3$ hrs ⟹ €90

Each step up costs another €20. Read down the fees:

$30, \; 50, \; 70, \; 90, \; 110, \; \ldots$

That ordered list is a sequence. And the equation tying it together is the line:

$y = 20x + 30$

Must learn. A sequence is an ordered list of numbers. Each value sits at a position $1, 2, 3, \ldots$

$\,T_1, \; T_2, \; T_3, \; \ldots, \; T_n, \; \ldots$

$T_n$ is the rule = general term = unknown term.

(i) Numbering the positions

For the plumber sequence $30, 50, 70, 90, \ldots$ we name the terms:

$T_1 = 30$

$T_2 = 50$

$T_3 = 70$

$T_n$ is whichever term we don't yet know.

Try this

For the sequence $5, \; 11, \; 17, \; 23, \; \ldots$ write down $T_1$, $T_2$, $T_3$ and $T_4$.

$T_1$ is the first one in the list. $T_2$ is the second. And so on.

$T_1 = 5$

$T_2 = 11$

$T_3 = 17$

$T_4 = 23$

$T_1=5,\; T_2=11,\; T_3=17,\; T_4=23$

Try this

A car costs €15,000 and loses €1,200 in value each year. Write the values for years $0, 1, 2, 3$ as $T_0, T_1, T_2, T_3$.

Subtract €1,200 from each previous value. Position $0$ is the start.

$T_0 = 15{,}000$

$T_1 = 15{,}000 - 1{,}200 = 13{,}800$

$T_2 = 13{,}800 - 1{,}200 = 12{,}600$

$T_3 = 11{,}400$

$15{,}000, \; 13{,}800, \; 12{,}600, \; 11{,}400$

Section 2

The X O □ repeating pattern

Look at this pattern:

X O □ X O □ X O □ $\ldots$

(a) What shape is in the $51^{\text{st}}$ position?

(b) Where is the $62^{\text{nd}}$ square?

(i) Spot the cycle

The pattern repeats every 3 positions. Label each shape's positions:

$T_1 = $ X $T_2 = $ O $T_3 = $ □

$T_4 = $ X $T_5 = $ O $T_6 = $ □

Each shape has its own rule:

□ at positions $3, 6, 9, 12, \ldots$ ⟹ $T_n = 3n$

O at positions $2, 5, 8, 11, \ldots$ ⟹ $T_n = 3n - 1$

X at positions $1, 4, 7, 10, \ldots$ ⟹ $T_n = 3n - 2$

(ii) What is the $51^{\text{st}}$?

Divide the position by $3$:

$\dfrac{51}{3} = 17$ (exact, no remainder)

Divisible by $3$ ⟹ the position lines up with the squares.

$T_{51} = $ □ (a square)

(iii) Where is the $62^{\text{nd}}$ square?

Squares sit at positions $T_n = 3n$. The $62^{\text{nd}}$ square is when $n = 62$.

Position $= 3(62)$

$= 186^{\text{th}}$ position

Try this

A pattern repeats: A B C A B C A B C $\ldots$ What letter is in the $47^{\text{th}}$ position?

Divide by $3$ and check the remainder. Remainder $0$ ⟹ C. Remainder $1$ ⟹ A. Remainder $2$ ⟹ B.

$\dfrac{47}{3} = 15$ remainder $2$

Remainder $2$ matches position $T_2 = $ B

$T_{47} = $ B

B

Try this

Same A B C pattern. What position is the $20^{\text{th}}$ B?

B sits at $T_2, T_5, T_8, \ldots$ — rule $T_n = 3n - 1$.

$T_n = 3n - 1$ for the B's

$20^{\text{th}}$ B ⟹ $n = 20$

Position $= 3(20) - 1$

$= 59^{\text{th}}$ position

$59^{\text{th}}$

Section 3

Linear (arithmetic) sequences

When the gap is the same between every consecutive pair, the sequence is linear — also called arithmetic.

Must learn. Linear sequence ⟺ same gap each step.

The rule looks like $y = mx + c$:
$\quad T_n = (\text{gap}) \cdot n + (\text{constant})$

(i) Carl's $T_0$ trick

To find the constant fast, step one back from $T_1$ and read off $T_0$.

Sequence: $\; 3, \; 7, \; 11, \; 15, \; \ldots$

Gap $= 4$

Step back: $\; T_0 = 3 - 4 = -1$

$T_n = 4n - 1$

Check: $T_1 = 4(1) - 1 = 3$ ✓

(ii) Find a far-out term

Once you have $T_n$, any term is one substitution away.

$T_{101} = 4(101) - 1$

$= 403$

(iii) A quick one

Sequence: $\; 2, \; 4, \; 6, \; 8, \; 10$

Gap $= 2$, so $T_0 = 0$

$T_n = 2n$

Try this

Find $T_n$ for the sequence $5, \; 8, \; 11, \; 14, \; \ldots$

Gap $= 3$. Step back from $5$ to find $T_0$.

Gap $= 3$

$T_0 = 5 - 3 = 2$

$T_n = 3n + 2$

Try this

Find $T_n$ for $\; 1, \; 6, \; 11, \; 16, \; \ldots$

Gap $= 5$. $T_0 = 1 - 5 = -4$.

Gap $= 5$

$T_0 = -4$

$T_n = 5n - 4$

Try this

Find $T_n$ for $\; 100, \; 95, \; 90, \; 85, \; \ldots$ (this one goes down)

Gap is negative. Gap $= -5$. Step back from $100$ adds $5$ to get $T_0$.

Gap $= -5$

$T_0 = 100 - (-5) = 105$

$T_n = -5n + 105$

$T_n = 105 - 5n$

Try this

For $\; 7, \; 10, \; 13, \; 16, \; \ldots$ find $T_{20}$.

Find $T_n$ first, then substitute $n = 20$.

Gap $= 3$, $T_0 = 4$

$T_n = 3n + 4$

$T_{20} = 3(20) + 4$

$= 64$

$T_{20} = 64$

Section 4

Computing terms from $T_n$

Going the other direction: given the rule $T_n$, substitute $n = 1, 2, 3, \ldots$ to get the actual numbers.

(i) Direct substitution

$T_n = 3n - 1$ — find the first three terms.

$T_1 = 3(1) - 1 = 2$

$T_2 = 3(2) - 1 = 5$

$T_3 = 3(3) - 1 = 8$

Try this

$T_n = 2n + 7$. Find $T_1$, $T_2$, $T_3$.

Plug $n = 1, 2, 3$ into the rule.

$T_1 = 2(1) + 7 = 9$

$T_2 = 2(2) + 7 = 11$

$T_3 = 2(3) + 7 = 13$

$9, \; 11, \; 13$

Try this

$T_n = n^2 - 1$. Find $T_1$, $T_3$, $T_5$.

Square the position, then subtract $1$. Watch the odd-only request.

$T_1 = 1 - 1 = 0$

$T_3 = 9 - 1 = 8$

$T_5 = 25 - 1 = 24$

$0, \; 8, \; 24$

Try this

$T_n = (-1)^n \cdot n$. Find the first four terms.

$(-1)^n$ flips sign each step. Odd $n$ ⟹ negative, even $n$ ⟹ positive.

$T_1 = (-1)(1) = -1$

$T_2 = (+1)(2) = 2$

$T_3 = (-1)(3) = -3$

$T_4 = (+1)(4) = 4$

$-1, \; 2, \; -3, \; 4$

Section 5

Algebraic manipulation of $T_n$

Notation. The terms either side of $T_n$ are:

$\; \ldots, \; T_{n-2}, \; T_{n-1}, \; T_n, \; T_{n+1}, \; T_{n+2}, \; \ldots$

$T_n$ is the general term — substitute $(n+1)$ or $(n-1)$ for $n$ to walk along the sequence algebraically.

(i) Finding $T_{n+1}$

$T_n = n^2$ — find $T_{n+1}$.

$T_{n+1} = (n+1)^2$

$= n^2 + 2n + 1$

(ii) An equation involving consecutive terms

Given $T_n = n^2$, solve $\; T_{n+1} - T_n = 57 \;$ for $n$.

$T_n = n^2$

$T_{n+1} = (n+1)^2 = n^2 + 2n + 1$

$T_{n+1} - T_n = (n^2 + 2n + 1) - n^2$

$\quad = 2n + 1$

$2n + 1 = 57$

$2n = 56$

$n = 28$

Sequence positions are natural numbers: $n \in \mathbb{N}$. If solving gives a negative or fractional root, throw it away.

Try this

$T_n = n^2 + 1$. Find $T_{n+1}$ in expanded form.

Replace every $n$ with $(n+1)$. Then expand the bracket.

$T_{n+1} = (n+1)^2 + 1$

$= n^2 + 2n + 1 + 1$

$= n^2 + 2n + 2$

$T_{n+1} = n^2 + 2n + 2$

Try this

$T_n = 3n - 2$. Show that $\; T_{n+1} - T_n = 3 \;$ for every $n$.

Write down $T_{n+1}$, then subtract $T_n$.

$T_{n+1} = 3(n+1) - 2 = 3n + 1$

$T_{n+1} - T_n = (3n + 1) - (3n - 2)$

$= 3$ ✓

$T_{n+1} - T_n = 3$. (This is why the gap of a linear sequence is constant.)

Try this

$T_n = n^2$. Solve $\; T_{n+1} - T_n = 21 \;$ for $n$, where $n \in \mathbb{N}$.

Same setup as the worked example. $T_{n+1} - T_n = 2n + 1$.

$T_{n+1} - T_n = 2n + 1$

$2n + 1 = 21$

$2n = 20$

$n = 10$

Section 6

Recurring relations

A recurring relation defines each term using one or more previous terms. You always need a starting value (or two).

Heads up — two notations.

$U_n = T_n$

Both letters mean the same thing. Exams use $T_n$ or $U_n$ interchangeably.

(i) One-step recurrence

$T_{n+1} = 2T_n$, with $T_1 = 5$. Find $T_4$.

$T_2 = 2T_1 = 2(5) = 10$

$T_3 = 2T_2 = 2(10) = 20$

$T_4 = 2T_3 = 2(20) = 40$

(ii) With a constant

$T_1 = 5$, $T_n = 2T_{n-1} + 3$. Find $T_3$.

$T_2 = 2T_1 + 3 = 2(5) + 3 = 13$

$T_3 = 2T_2 + 3 = 2(13) + 3 = 29$

(iii) Fibonacci-style

A two-back recurrence needs two starting values.

$U_0 = U_1 = 1$, $U_{n+2} = U_{n+1} + U_n$. Find $U_5$.

$U_2 = U_1 + U_0 = 1 + 1 = 2$

$U_3 = U_2 + U_1 = 2 + 1 = 3$

$U_4 = U_3 + U_2 = 3 + 2 = 5$

$U_5 = U_4 + U_3 = 5 + 3 = 8$

Sequence: $\; 1, \; 1, \; 2, \; 3, \; 5, \; 8, \; 13, \; \ldots$ — the famous Fibonacci sequence.

Try this

$T_1 = 3$, $T_{n+1} = 4T_n - 2$. Find $T_3$.

Use $T_1$ to get $T_2$, then $T_2$ to get $T_3$.

$T_2 = 4T_1 - 2 = 4(3) - 2 = 10$

$T_3 = 4T_2 - 2 = 4(10) - 2 = 38$

$T_3 = 38$

Try this

$U_0 = 1$, $U_1 = 2$, $U_{n+2} = U_{n+1} + 2U_n$. Find $U_4$.

Two-back recurrence. Need $U_2$ first, then $U_3$, then $U_4$.

$U_2 = U_1 + 2U_0 = 2 + 2(1) = 4$

$U_3 = U_2 + 2U_1 = 4 + 2(2) = 8$

$U_4 = U_3 + 2U_2 = 8 + 2(4) = 16$

$U_4 = 16$

Try this

$T_1 = 0$, $T_{n+1} = T_n + n$. Find $T_5$.

The amount you add changes each step — it's $n$, not a constant. Going from $T_1$ to $T_2$ uses $n=1$.

$T_2 = T_1 + 1 = 0 + 1 = 1$

$T_3 = T_2 + 2 = 1 + 2 = 3$

$T_4 = T_3 + 3 = 3 + 3 = 6$

$T_5 = T_4 + 4 = 6 + 4 = 10$

$T_5 = 10$ (the triangular numbers $0, 1, 3, 6, 10, \ldots$)

Try this

$T_1 = 1$, $T_n = 3T_{n-1} + 1$. Find $T_4$.

Apply $\times 3$ then $+ 1$ three times in a row.

$T_2 = 3(1) + 1 = 4$

$T_3 = 3(4) + 1 = 13$

$T_4 = 3(13) + 1 = 40$

$T_4 = 40$

Section 7

Proving a recurrence from $T_n$

Sometimes you're given a closed form $U_n$ and asked to prove it satisfies a recurrence. The standard exam trick: split $U_n$ into two pieces $a$ and $b$, then work out $U_{n+1}$ and $U_{n+2}$ in terms of $a$ and $b$.

Must learn. When $U_n = p^n + q^n$, write

$a = p^n, \quad b = q^n$

Uses the index law $\; a^{p+q} = a^p \cdot a^q$.

(i) Worked example

$U_n = 2^n + 3^n$. Prove $\; U_{n+2} - 5U_{n+1} + 6U_n = 0$.

Let $\; a = 2^n, \; b = 3^n$. So $U_n = a + b$.

$U_{n+1} = 2^{n+1} + 3^{n+1}$

$\quad\quad = 2 \cdot 2^n + 3 \cdot 3^n$

$\quad\quad = 2a + 3b$

$U_{n+2} = 2^{n+2} + 3^{n+2}$

$\quad\quad = 4 \cdot 2^n + 9 \cdot 3^n$

$\quad\quad = 4a + 9b$

$U_{n+2} - 5U_{n+1} + 6U_n$

$\quad = (4a + 9b) - 5(2a + 3b) + 6(a + b)$

$\quad = 4a + 9b - 10a - 15b + 6a + 6b$

$\quad = (4 - 10 + 6)a + (9 - 15 + 6)b$

$\quad = 0 \cdot a + 0 \cdot b = 0$ ✓

Try this

$U_n = 5^n + 7^n$. Prove $\; U_{n+2} - 12U_{n+1} + 35U_n = 0$.

Let $a = 5^n$, $b = 7^n$. Coefficients you'll need: $5, 7, 25, 49$.

Let $a = 5^n$, $b = 7^n$. So $U_n = a + b$.

$U_{n+1} = 5 \cdot 5^n + 7 \cdot 7^n = 5a + 7b$

$U_{n+2} = 25 \cdot 5^n + 49 \cdot 7^n = 25a + 49b$

$U_{n+2} - 12U_{n+1} + 35U_n$

$= 25a + 49b - 12(5a + 7b) + 35(a + b)$

$= 25a + 49b - 60a - 84b + 35a + 35b$

$= (25 - 60 + 35)a + (49 - 84 + 35)b$

$= 0$ ✓

Coefficients of both $a$ and $b$ collapse to $0$ ⟹ result is $0$.

Try this

$U_n = 3^n + 4^n$. Prove $\; U_{n+2} - 7U_{n+1} + 12U_n = 0$.

Let $a = 3^n$, $b = 4^n$. Coefficients: $3, 4, 9, 16$. The numbers $7$ and $12$ are $3+4$ and $3 \times 4$ — not a coincidence.

Let $a = 3^n$, $b = 4^n$. So $U_n = a + b$.

$U_{n+1} = 3a + 4b$

$U_{n+2} = 9a + 16b$

$U_{n+2} - 7U_{n+1} + 12U_n$

$= 9a + 16b - 7(3a + 4b) + 12(a + b)$

$= 9a + 16b - 21a - 28b + 12a + 12b$

$= (9 - 21 + 12)a + (16 - 28 + 12)b$

$= 0$ ✓

$= 0$ ✓ (The recurrence comes from $(x-3)(x-4) = x^2 - 7x + 12$.)

Section 8

Quadratic sequences (gap of gap)

When the gap isn't constant, take the gap of the gap. If that's constant, the sequence is quadratic.

Must learn.

Gap of gap constant ⟹ $T_n = an^2 + bn + c$

And the coefficient of $n^2$ comes straight off the second difference:

$\; a = \dfrac{\text{gap of gap}}{2}$

(i) Worked example

Find $T_n$ for $\; 11, \; 18, \; 29, \; 44, \; 63$.

Gaps: $\quad 7, \quad 11, \quad 15, \quad 19$

Gap of gap: $\quad 4, \quad 4, \quad 4$

$a = \dfrac{4}{2} = 2$

$T_n = 2n^2 + bn + c$

$T_1 = 11$: $2(1)^2 + b + c = 11$ ⟹ $b + c = 9$

$T_2 = 18$: $2(2)^2 + 2b + c = 18$ ⟹ $2b + c = 10$

Subtract the equations:

$b = 1$

Then $c = 9 - 1 = 8$

$T_n = 2n^2 + n + 8$

Check: $T_3 = 2(9) + 3 + 8 = 29$ ✓

(ii) Triangular numbers

Find $T_n$ for $\; 1, \; 3, \; 6, \; 10$.

Gaps: $\quad 2, \quad 3, \quad 4$

Gap of gap: $\quad 1, \quad 1$

$a = \dfrac{1}{2}$

$T_n = \dfrac{1}{2}n^2 + bn + c$

$T_1 = 1$: $\dfrac{1}{2} + b + c = 1$ ⟹ $2b + 2c = 1$

$T_2 = 3$: $\dfrac{1}{2}(4) + 2b + c = 3$ ⟹ $2b + c = 1$

Subtract: $c = 0$

Then: $b = \dfrac{1}{2}$

$T_n = \dfrac{1}{2}n^2 + \dfrac{1}{2}n$

Check: $T_3 = \dfrac{1}{2}(9) + \dfrac{1}{2}(3) = 4.5 + 1.5 = 6$ ✓

(iii) The $T_0$ shortcut

Sub $n = 0$ into $T_n = an^2 + bn + c$. You get $T_0 = c$. So $c$ can be read straight off the term before $T_1$.

For $1, 3, 6, 10, \ldots$ extend one back:

$\; T_0 = 1 - 2 = 0, \; T_1 = 1, \; T_2 = 3, \; \ldots$

So $c = T_0 = 0$ (matches what we calculated).

Try this

Find $T_n$ for $\; 4, \; 9, \; 16, \; 25, \; 36$.

Gaps: $5, 7, 9, 11$. Gap of gap: $2$. So $a = 1$.

Gaps: $5, 7, 9, 11$; gap of gap $= 2$; $a = 1$

$T_n = n^2 + bn + c$

$T_1 = 4$: $1 + b + c = 4$ ⟹ $b + c = 3$

$T_2 = 9$: $4 + 2b + c = 9$ ⟹ $2b + c = 5$

Subtract: $b = 2$, then $c = 1$

$T_n = n^2 + 2n + 1 = (n+1)^2$

$T_n = (n+1)^2$

Try this

Find $T_n$ for $\; 3, \; 8, \; 15, \; 24, \; 35$.

Gaps $5, 7, 9, 11$. Gap of gap $2$. $a = 1$.

Gaps: $5, 7, 9, 11$; gap of gap $= 2$; $a = 1$

$T_n = n^2 + bn + c$

$T_1 = 3$: $b + c = 2$

$T_2 = 8$: $2b + c = 4$

Subtract: $b = 2$, $c = 0$

$T_n = n^2 + 2n$

$T_n = n^2 + 2n$ or equivalently $n(n+2)$

Try this

Find $T_n$ for $\; 2, \; 5, \; 10, \; 17, \; 26$.

Try the $T_0$ shortcut: extend back one — gap of gap is $2$, so $a = 1$ and you can write $T_0$ first.

Gaps: $3, 5, 7, 9$; gap of gap $= 2$; $a = 1$

$T_n = n^2 + bn + c$

$T_1 = 2$: $b + c = 1$

$T_2 = 5$: $2b + c = 1$

Subtract: $b = 0$, $c = 1$

$T_n = n^2 + 1$

Try this

Find $T_n$ for $\; 8, \; 11, \; 16, \; 23, \; 32$.

Gaps $3, 5, 7, 9$. Gap of gap $2$. $a = 1$. Notice $T_0$ trick: step back gives $T_0 = 8 - 3 = 5$? No — the first gap isn't yet constant. Solve with $T_1, T_2$ instead.

Gaps: $3, 5, 7, 9$; gap of gap $= 2$; $a = 1$

$T_n = n^2 + bn + c$

$T_1 = 8$: $b + c = 7$

$T_2 = 11$: $2b + c = 7$

Subtract: $b = 0$, $c = 7$

$T_n = n^2 + 7$

Section 9

Increasing or decreasing

Some sequences only ever go up. Others only ever go down.

$2, \; 4, \; 6, \; 8, \; \ldots$ increasing

$\dfrac{1}{2}, \; \dfrac{1}{3}, \; \dfrac{1}{4}, \; \dfrac{1}{5}, \; \ldots$ decreasing

To prove which one a given $T_n$ is, compare $T_{n+1}$ with $T_n$. The cleanest tool for fractions and exponentials is the ratio test.

Must learn. For positive terms:

$\dfrac{T_{n+1}}{T_n} > 1$ ⟹ increasing

$\dfrac{T_{n+1}}{T_n} < 1$ ⟹ decreasing

Divide rather than subtract — for fractions and exponentials, the cancellation is much cleaner.

(i) Worked example

Is $\; T_n = \dfrac{n+1}{n-1} \;$ increasing?

$T_{n+1} = \dfrac{(n+1)+1}{(n+1)-1} = \dfrac{n+2}{n}$

$\dfrac{T_{n+1}}{T_n} = \dfrac{n+2}{n} \cdot \dfrac{n-1}{n+1}$

$\quad\quad = \dfrac{n^2 + n - 2}{n^2 + n}$

Top vs bottom: the top is $(n^2 + n) - 2$, which is $2$ less than the bottom.

$\dfrac{n^2 + n - 2}{n^2 + n} < 1$

⟹ decreasing

Carl's catchphrase: "Tip is less than bottom." If the numerator is smaller, the fraction is less than $1$.

(ii) Same idea, harder algebra

Is $\; T_n = \dfrac{2n+1}{2n-1} \;$ increasing?

$T_{n+1} = \dfrac{2(n+1)+1}{2(n+1)-1} = \dfrac{2n+3}{2n+1}$

$\dfrac{T_{n+1}}{T_n} = \dfrac{2n+3}{2n+1} \cdot \dfrac{2n-1}{2n+1}$

$\quad\quad = \dfrac{(2n+3)(2n-1)}{(2n+1)^2}$

$\quad\quad = \dfrac{4n^2 + 4n - 3}{4n^2 + 4n + 1}$

Top is bottom $-4$. So top is less than bottom.

$\dfrac{4n^2 + 4n - 3}{4n^2 + 4n + 1} < 1$

⟹ decreasing

Try this

Is $\; T_n = \dfrac{3n+1}{3n-1} \;$ increasing or decreasing?

Same shape as the worked example. Get $T_{n+1}$ by replacing $n$ with $n+1$. Take the ratio and compare top with bottom.

$T_{n+1} = \dfrac{3(n+1)+1}{3(n+1)-1} = \dfrac{3n+4}{3n+2}$

$\dfrac{T_{n+1}}{T_n} = \dfrac{3n+4}{3n+2} \cdot \dfrac{3n-1}{3n+1}$

$\quad = \dfrac{(3n+4)(3n-1)}{(3n+2)(3n+1)} = \dfrac{9n^2 + 9n - 4}{9n^2 + 9n + 2}$

Top $-$ bottom $= -6 < 0$, so tip $<$ bottom.

Ratio $< 1$ ⟹ decreasing

Decreasing

Try this

Is $\; T_n = \dfrac{n}{n+1} \;$ increasing or decreasing?

$T_1 = \dfrac{1}{2}$, $T_2 = \dfrac{2}{3}$, $T_3 = \dfrac{3}{4}$ — going up? Confirm with the ratio test.

$T_{n+1} = \dfrac{n+1}{n+2}$

$\dfrac{T_{n+1}}{T_n} = \dfrac{n+1}{n+2} \cdot \dfrac{n+1}{n} = \dfrac{(n+1)^2}{n(n+2)}$

$\quad = \dfrac{n^2 + 2n + 1}{n^2 + 2n}$

Tip is bottom $+1$ ⟹ ratio $> 1$.

Increasing

Try this

Is $\; T_n = \dfrac{n+5}{n+2} \;$ increasing or decreasing?

$T_1 = \dfrac{6}{3} = 2$, $T_2 = \dfrac{7}{4} = 1.75$. So it's going down — confirm with the ratio test.

$T_{n+1} = \dfrac{n+6}{n+3}$

$\dfrac{T_{n+1}}{T_n} = \dfrac{n+6}{n+3} \cdot \dfrac{n+2}{n+5} = \dfrac{(n+6)(n+2)}{(n+3)(n+5)}$

$\quad = \dfrac{n^2 + 8n + 12}{n^2 + 8n + 15}$

Tip is bottom $-3$ ⟹ ratio $< 1$.

Decreasing

That's Sequences & Series — Part 1.

You've covered patterns, the general term $T_n$, linear & recurring relations, proving recurrences, quadratic sequences and increasing/decreasing. Part 2 will pick up with $\Sigma$ notation, arithmetic series and geometric sequences.