Ms6–Ms11 - Sequences, recurrence and sigma notation

You already know

Substitution puts an input into a rule. In $3x-2$, replace $x$ by the chosen input, multiply by $3$, then subtract $2$.

Sequence and recurrence

A sequence is an ordered list of numbers, with a term at each allowed integer position. A kth-term rule gives a term directly from its position, such as $a_k=2+3k$. A recurrence gives the next term from an earlier one, such as $a_{k+1}=3a_k-2$, and needs initial information. State whether indexing starts at $0$ or $1$.

Worked example 1: recurrence values are the inputs

Question. Given $a_1=2$ and $a_{k+1}=3a_k-2$, find $a_2,a_3,a_4$.

Plan. Feed each newly calculated value into the same rule.

Why not put the index $k$ into the rule? The recurrence requires $a_k$, the term value, as its input.

Answer. $4,10,28$.

Check. The differences from $1$ are $1,3,9,27$, each three times the previous difference, consistent with $a_{k+1}-1=3(a_k-1)$.

Worked example 2: a fixed starting term

Question. Given $a_1=1$ and $a_{k+1}=3a_k-2$, find the first four terms and explain what happens afterwards.

Plan. Test whether the recurrence returns $1$ unchanged.

Why can we describe all later terms? Once the input is $1$, every subsequent output is exactly $1$.

Answer. The sequence is $1,1,1,1,\ldots$, so it is constant.

Check. Solving $L=3L-2$ gives the fixed value $L=1$.

Watch out

Wrong: a recurrence alone specifies one sequence. Right: its initial value matters, as the two worked examples show. Do not confuse a position rule with a term-to-term rule.

Increasing, decreasing and periodic

Here increasing means $a_{k+1}>a_k$ for each consecutive pair; decreasing means $a_{k+1}<a_k$. A periodic sequence satisfies $a_{k+p}=a_k$ for all permitted $k$, for some fixed positive integer $p$. Its period is the smallest such $p$. A constant sequence has period $1$, and is neither strictly increasing nor strictly decreasing.

Worked example 3: establish a repeating block

Question. For $c_1=1$, $c_{k+1}=6-c_k$, find the first four terms and describe the sequence's behaviour.

Plan. Generate terms, then apply the recurrence twice symbolically.

Why is the symbolic step useful? It proves that the observed block repeats indefinitely.

Answer. It is infinite, periodic with period $2$, and neither increasing nor decreasing.

Check. The two consecutive values differ, so period $1$ is impossible.

Worked example 4: a shorter repeating block

Question. For $c_1=3$, $c_{k+1}=6-c_k$, describe the sequence and its smallest period.

Plan. Check whether a single application fixes the initial value.

Why not say the period is $2$? Two steps do return the input, but one step already does so.

Answer. $3,3,3,\ldots$ is infinite and constant, with period $1$; it is neither strictly increasing nor strictly decreasing.

Check. Every difference $c_{k+1}-c_k$ equals $0$.

Watch out

Wrong: “a bounded sequence is finite”. Right: bounded describes the size of its terms; finite describes how many positions it has. An endless two-value cycle is bounded and infinite.

Partial sums and sigma notation

Sigma instructs you to add the term expression over every integer index from the lower limit to the upper limit, both included. The index is a dummy variable: changing its letter consistently leaves the sum unchanged.

Not in the formula sheet: learn it.