Read terms and check limits

Track the starting index, distinguish u_n from S_n, and test convergence before taking an infinite limit. Consecutive partial sums recover a term: $u_n=S_n-S_{n-1}$.

Read terms, positions and partial sums

A sequence is an ordered list. The source gives the finite lists $3,4,5,6,7$ and $1,1/2,1/3,1/4,1/5$, and the infinite lists $2,4,6,8,\ldots$ and $1,1,2,3,5,8,13,21,\ldots$. In the first list, the first term is 3, the second 4 and the fifth and last 7. An ellipsis indicates continuing terms; a finite sequence has a last term. The source links Fibonacci music through its QR code.

In $u_1,u_2,u_3,\ldots,u_n,\ldots$, the subscript is the index, meaning position. Here $n$ is a positive integer, as the Teacher emphasises. Adding terms gives a series; its partial sum is

$$S_n=u_1+u_2+\cdots+u_n.$$

An explicit formula $u_n=f(n)$ takes the position as input and produces that term. A finite sequence has a finite domain such as $\{1,2,3,4,5\}$; an infinite sequence can have domain $\mathbb Z^+$. This qualifies the source’s Function callout: a finite list does not have a term at every positive index. For $u_n=n^2+3$, substituting $n=1,2,3,4,5$ gives $4,7,12,19,28$, and $u_{100}=100^2+3=10003$. Self-Practice 1 checks both substitution and choosing a formula with the correct domain.

The supplied calculator instructions use the device shown in the source:

1. Press MODE and choose SEQ.
2. Press y=, enter $u(n)=n^2+3$ using the X,T,θ,n key for $n$, and set nMin=1. The explicit formula does not need the $u(1),u(2)$ initial-value fields filled.
3. Press 2nd graph to view the table, using Up/Down to inspect indices. The displayed index 0 gives ERROR because of the stated sequence domain; the polynomial itself is mathematically evaluable at 0, but that would extend this sequence’s domain.
4. Press 2nd window and set TblStart=100 rather than scrolling from the beginning.
5. Press 2nd graph again. The source’s table 100–110 gives $10003,10204,10407,10612,10819,11028,11239,11452,11667,11884,12103$.

Its first table shows indices 1–10 with $4,7,12,19,28,39,52,67,84,103$. These finite tables illustrate increase. Algebra proves it for all positive indices: $u_{n+1}-u_n=2n+1>0$; the values also grow without bound. A calculator table is evidence for a conjecture, not a proof about infinitely many terms.

The source outline sets out the later chapter route: finite/infinite sequences and series; sequences as positive-integer-input functions; the relation between $u_n$ and $S_n$; explicit formulas; recurrences $u_{n+1}=f(u_n)$; sigma notation; sums/differences of series; convergence and infinite sums; arithmetic nth-term/finite-sum formulas; geometric nth-term/finite-sum formulas; geometric convergence conditions; and the convergent geometric infinite-sum formula. The opening material here does not replace those later source sections.

Separate behaviour from convergence

Ask two independent questions: does each term rise or fall, and do the terms approach one finite value? In the source’s strict terminology, increasing means $u_{n+1}>u_n$ and decreasing means $u_{n+1}<u_n$. A sequence may be neither.

Write

$$\lim_{n\to\infty}u_n=L$$

when the terms approach one finite constant $L$. Read lim as “the limiting value of”, the arrow as “approaches” or “tends to”, $u_n$ as the sequence formula, and $L$ as the limiting value.

The complete statement reads “the limit of the sequence $u_n$ as $n$ tends to infinity is $L$”. This notation reappears in differentiation.

A sequence is convergent when such a finite limit exists and divergent otherwise; $u_n\to+\infty$ describes divergence, not a finite limit.

The source’s Limits callout also discusses a series’ limiting sum. Keep the objects separate: sequence convergence concerns $u_n$, while convergence of a series concerns the sequence of partial sums $S_n$. A sequence’s terms tending to zero alone does not establish that their series has a finite sum.

Both printed lists for $-n^2$ skip $-16$; it is restored above from the formula. The printed “$w_n\to-1$ or $1$” describes the two alternating values, not one sequence limit. Its odd and even subsequences have different limits, so the whole sequence diverges.

Example 1 shows why the two questions must stay separate. The alternation $(-1/2)^n$ converges to 0 because its magnitude tends to 0, while $(-2)^n$ diverges because its magnitudes grow without bound.

The sequence $2+e^{-n}$ decreases to 2. Its exact terms are greater than 2 for every finite positive index, even when the calculator displays 2 after rounding.

The Teacher asks you to notice this limitation.

Likewise, the displayed starting index 100 is an arbitrary sample, not a mathematical infinity or cutoff.

Apply limit laws with their hypotheses

If $u_n\to A$ and $v_n\to B$ with finite $A,B$, the source laws are

$$\lim_{n\to\infty}(u_n\pm v_n)=A\pm B,\qquad \lim_{n\to\infty}(c u_n)=cA$$

for a constant $c$, and

$$\lim_{n\to\infty}\frac{u_n}{v_n}=\frac AB\quad\text{provided }B\ne0.$$

Check these hypotheses before substituting limits. The source warns against treating $0/\infty$, $\infty/\infty$ or $0/0$ as ordinary real-number divisions. Infinity is not a real number. In limit analysis, a numerator tending to 0 over a denominator whose magnitude tends to infinity does tend to 0; the warning does not forbid that conclusion. The forms $0/0$ and $\infty/\infty$ alone are inconclusive and require rewriting the sequence.

Rewrite rational formulas before taking limits

Example 2 deliberately forbids a calculator. Divide numerator and denominator by a suitable power of $n$, or use long division to expose a polynomial plus a proper rational remainder. For positive $n$, $3n-1$ is nonzero.

In(a), expanding the squared denominator and dividing by $n^2$ gives

$$\frac{2n+1}{(3n-1)^2} =\frac{2/n+1/n^2}{9-6/n+1/n^2}\to0.$$

In(b), both printed methods give the same finite limit:

$$\frac{2n+1}{3n-1}=\frac23+\frac5{3(3n-1)} =\frac{2+1/n}{3-1/n}\to\frac23.$$

In(c), long division gives

$$\frac{6n^2+n+1}{3n-1}=2n+1+\frac2{3n-1}.$$

The proper rational remainder tends to 0, but $2n+1$ grows without bound, so the whole sequence diverges to $+\infty$. The full solutions retain every printed intermediate step and both methods for (b).

Generate a sequence from earlier terms

A recurrence relation expresses a term using one or more previous terms, together with enough initial values to start. The source shows $u_{n+1}=f(u_n)$ as a common form. Recurrences can also depend explicitly on $n$ or on two preceding terms, as its examples demonstrate. An explicit formula takes the index as input directly; a recurrence advances through earlier terms.

The three source examples pair a recurrence with an explicit formula:

• Squares: $u_{n+1}=u_n+2n+1$, $u_1=1$, gives $1,4,9,16,25,36,\ldots$ and $u_n=n^2$. Indeed, $(n+1)^2-n^2=2n+1$.
• Harmonic sequence: $1/u_{n+1}=1/u_n+1$, $u_1=1$, gives $1,1/2,1/3,1/4,1/5,\ldots$ and $u_n=1/n$. The reciprocals increase by one. Its terms are nonzero; this name describes the sequence, not a claim that its infinite series has a finite sum.
• Fibonacci sequence: $u_n=u_{n-1}+u_{n-2}$ for $n\ge3$, with $u_1=u_2=1$, gives $1,1,2,3,5,8,13,21,34,\ldots$. Two initial terms are needed here. Its printed explicit formula is

$$u_n=\frac1{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right].$$

The source gives this formula without deriving it. The recurrence supplies a simple way to generate its integer terms without approximating the square roots.

Match the calculator setup to the recurrence

Example 3 uses $u_n=2u_{n-1}+3$, $u_1=2$. Start at the given first term and apply the relation from $n=2$ onwards. On the device shown in the source:

1. Press MODE and choose SEQ.
2. Press y=, select the current-term form SEQ(n) and set nMin=1.
3. Press 2nd 7 for the $u$ function and enter $2u(n-1)+3$; use X,T,θ,n for $n$.
4. Set u(1)=2. The source leaves u(2) blank because the recurrence computes it from the given first term.
5. Press 2nd graph and inspect the table with Up/Down. Its first eleven terms are $2,7,17,37,77,157,317,637,1277,2557,5117$.

The requested $u_{20}=2621437$ comes after nineteen updates. The worked solution also gives an independent shifted-form check; that algebra supplements the printed calculator method.

Example 4 starts with $u_{n+1}-3u_n=4$, $u_1=5$. Isolate the next term first:

$$u_{n+1}=3u_n+4.$$

For the source’s second method, keep the calculator in SEQ mode, press y=, use Up/Right to highlight TYPE SEQ(n+1) and press Enter. Set nMin=1; use 2nd 7 for $u$, X,T,θ,n for $n$, enter $3u(n)+4$, and set u(1)=5. 2nd graph with Up/Down gives $5,19,61,187,565,1699,5101,15307,45925,137779,413341$ at indices 1–11. Its requested first three new terms agree with direct substitution. The recurrence notation and the selected calculator type must describe the same update.