Precalculus Sequences & Series Guide | Engineering Confidence

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BEFORE YOU COMPUTE

About this review. The review packet has 21 problems in flashcard format — each page shows the previous answer above the next problem, so the topics jump around. That’s great for studying, because it forces you to do the hardest step first — tool selection — on every problem.

This guide presents the 21 problems in the same order as the printed review packet, so you can work alongside it. Each header names the concept the problem tests. The next two pages are your concept reference: a diagnostic decision tree and a Master Toolbox. Read those first, then work through the problems and refer back as needed.

Diagnostic Decision Tree

HOW TO READ THE PROMPT

When you sit down with a problem, run through these questions in order. Stop at the first “yes.”

1. Is it a sequence? (one term, or the formula for $a_{n}$ )

Look at numerators, denominators, and signs separately.
Test your formula by plugging in $n = 1, 2, 3$ and matching given terms.
Tools: arithmetic ( $a_{n} = a_{1} + (n - 1) d$ ), geometric ( $a_{n} = a_{1} r^{n - 1}$ ), or alternating signs via ${(- 1)}^{n}$ or ${(- 1)}^{n + 1}$ .

2. Is it a finite sum (something like $1 + 4 + 7 + \dots + 295$ , or $\sum_{n = 1}^{8} \dots$ )?

Constant difference between terms? Arithmetic. Use $S_{n} = \frac{n}{2} (a_{1} + a_{n})$ .
Constant ratio between terms? Geometric. Use $S_{n} = a_{1} \cdot \frac{1 - r^{n}}{1 - r}$ .

3. Is it an infinite sum (upper limit is $\infty$ )? Run the convergence pipeline:

(a): Geometric with $| r | < 1$ ? Use $S = \frac{a_{1}}{1 - r}$ .
(b): Test for Divergence: compute $\lim_{n \to \infty} a_{n}$ . If it isn’t $0$ , write diverges and stop.
(c): Direct Comparison: bound your terms above by a known convergent series, or below by a known divergent one (often the harmonic $\sum 1 ∕ n$ ).

4. Is it a repeating decimal? Express the repeating tail as an infinite geometric series. First repeated block at its actual decimal location = $a_{1}$ ; ratio = $1 0^{- k}$ where $k$ is the block length. Apply $S = \frac{a_{1}}{1 - r}$ , then add the non-repeating prefix.

5. Is there an $x$ inside the series? (“Find the interval of convergence.”) Treat the bracketed $x$ -expression as the ratio of a geometric series. Set $| bracket | < 1$ and solve for $x$ .

6. Does the series have factorials in the denominator? Probably a Taylor series. Match against the three you should have memorized:

$e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$
$\sin x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}$
$\cos x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!}$

Master Toolbox — All the Formulas You Need

SEQUENCES

Arithmetic: common difference $d$ .

$n$ th term: $a_{n} = a_{1} + (n - 1) d$
Recursive form: $a_{n} = a_{n - 1} + d$

Geometric: common ratio $r$ .

$n$ th term: $a_{n} = a_{1} \cdot r^{n - 1}$
Recursive form: $a_{n} = r \cdot a_{n - 1}$

Identifying which one: subtract consecutive terms (constant $\Rightarrow$ arithmetic) or divide consecutive terms (constant $\Rightarrow$ geometric).

SERIES SUMS

Arithmetic: $S_{n} = \frac{n}{2} (a_{1} + a_{n}) = \frac{n}{2} [2 a_{1} + (n - 1) d]$

Geometric (finite): $S_{n} = a_{1} \cdot \frac{1 - r^{n}}{1 - r}$ (use whenever $r \neq 1$ )

Geometric (infinite): $S = \frac{a_{1}}{1 - r}$ (only if $| r | < 1$ — otherwise the sum diverges)

Sigma-notation arithmetic: $\sum_{n = 1}^{N} c = 𝑐𝑁$ , $\sum_{n = 1}^{N} n = \frac{N (N + 1)}{2}$

CONVERGENCE TESTS FOR $\sum_{n = 1}^{\infty} a_{n}$

1.

Geometric series test. If

a_{n} = a_{1} r^{n - 1}

, the series converges iff

| r | < 1

2.

Test for Divergence (TfD). If

\lim_{n \to \infty} a_{n} \neq 0

(including doesn’t exist), the series diverges. If the limit is

0

, the test is inconclusive — it does not prove convergence.

3.

Direct Comparison. Suppose

0 \leq a_{n} \leq b_{n}

for all

n

past some point.

If $\sum b_{n}$ converges, then $\sum a_{n}$ converges (smaller than something finite).
If $\sum a_{n}$ (the smaller one) diverges, then $\sum b_{n}$ also diverges.

The harmonic series $\sum_{n = 1}^{\infty} \frac{1}{n}$ diverges — this is your reference for “terms going to $0$ but the sum still blowing up.”

LIMIT-OF- $a_{n}$ SHORTCUTS

You’ll use these inside the Test for Divergence and inside Taylor recognition:

Rational $a_{n} = \frac{P (n)}{Q (n)}$ : compare degrees of top and bottom.
- Top deg $<$ bottom deg $\Rightarrow$ limit $= 0$ .
- Top deg $=$ bottom deg $\Rightarrow$ limit $=$ ratio of leading coefficients.
- Top deg $>$ bottom deg $\Rightarrow$ limit $= \pm \infty$ .
Geometric $r^{n}$ : $\to 0$ if $| r | < 1$ ; $\to + \infty$ if $r > 1$ ; no limit (oscillates with growing magnitude) if $r < - 1$ ; oscillates if $r = - 1$ ; $\to 1$ if $r = 1$ .
“Drop the small terms.” For limits at infinity, only the dominant term in numerator and denominator matters. The $+ 5$ in $4 n + 5$ , the $+ 21$ in $73 n + 21$ , and the ${(- 1)}^{n}$ in $37 n + {(- 1)}^{n}$ are all bounded $\Rightarrow$ ignore them when computing the limit.

TAYLOR SERIES — MEMORIZE THESE THREE

For all real $x$ :

e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}, \sin x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}, \cos x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!} .

Pattern recognition cues:

Factorial $n!$ in denominator and $x^{n}$ on top $\Rightarrow$ $e^{x}$ .
Alternating sign, factorial odd $(2 n + 1)!$ , odd power $x^{2 n + 1}$ $\Rightarrow$ $\sin x$ .
Alternating sign, factorial even $(2 n)!$ , even power $x^{2 n}$ $\Rightarrow$ $\cos x$ .

Bonus geometric: for $| x | < 1$ , $\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}$ . (This is just the infinite-geometric-series formula written as a function of $x$ .)

The 21 Problems — In Review-Packet Order

The problems appear below in the same order they’re given in the printed review packet. Each header notes the concept the problem tests — the Master Toolbox on the previous pages is your concept reference. Work the problem first, then read the scaffolding.

PROBLEM 1

repeating decimal with non-repeating prefix

Use a geometric series to write $- 5.8 \bar{67}$ as a ratio of integers.

BEFORE YOU COMPUTE

Reading the decimal. The bar is over “ $67$ ” only, so $- 5.8 \bar{67} = - 5.86767676 \dots$ . The “ $8$ ” is part of a fixed (non-repeating) prefix; the “ $67$ ” is the repeating block.

The question explicitly says “use a geometric series.” So our job is to express the repeating tail as $\sum_{n = 0}^{\infty} a_{1} r^{n}$ and apply the infinite-geometric-sum formula $S = \frac{a_{1}}{1 - r}$ .

Strategy.

1.: Pull the negative sign out front; deal with $5.8 \bar{67}$ as a positive number, then re-apply the sign at the end.
2.: Split into non-repeating + repeating: $5.8 \bar{67} = 5.8 + 0.0 \bar{67}$ .
3.: Write the repeating tail $0.0 \bar{67}$ as an infinite geometric series in sigma notation.
4.: Identify $a_{1}$ and $r$ , check $| r | < 1$ , apply $S = \frac{a_{1}}{1 - r}$ .
5.: Add to $5.8$ , re-apply the negative.

WORKING

Step 1: Split the decimal. The non-repeating part is $5.8 = \frac{58}{10}$ , and the repeating tail is $0.0 \bar{67} = 0.0676767 \dots$ :

5.8 \bar{67} = 5.8 + 0.0 \bar{67} .

Step 2: Write the repeating tail as a sum of fractions. The first “ $67$ ” lives in positions $1 0^{- 2}$ and $1 0^{- 3}$ :

0.067 = \frac{67}{1000} = \frac{67}{1 0^{3}} .

The next “ $67$ ” is shifted two more decimal places to the right (since the block has length $2$ ), so it contributes $\frac{67}{1 0^{5}}$ . Then $\frac{67}{1 0^{7}}$ , then $\frac{67}{1 0^{9}}$ , and so on:

0.0 \bar{67} = \frac{67}{1 0^{3}} + \frac{67}{1 0^{5}} + \frac{67}{1 0^{7}} + \frac{67}{1 0^{9}} + \dots

Step 3: Put in sigma notation and identify the geometric pieces. Pull out the common factor $\frac{67}{1 0^{3}}$ :

0.0 \bar{67} = \frac{67}{1 0^{3}} (1 + \frac{1}{1 0^{2}} + \frac{1}{1 0^{4}} + \frac{1}{1 0^{6}} + \dots) = \frac{67}{1000} \sum_{n = 0}^{\infty} {(\frac{1}{100})}^{n} .

This is an infinite geometric series with

a_{1} = \frac{67}{1000}, r = \frac{1}{100} .

Convergence check: $| r | = \frac{1}{100} < 1$ ✓, so the sum exists.

Step 4: Apply $S = \frac{a_{1}}{1 - r}$ .

0.0 \bar{67} = \frac{67 ∕ 1000}{1 - 1 ∕ 100} = \frac{67 ∕ 1000}{99 ∕ 100} = \frac{67}{1000} \cdot \frac{100}{99} = \frac{67}{990} .

Step 5: Combine with the non-repeating part. Use the common denominator $990$ :

5.8 = \frac{58}{10} = \frac{58 \cdot 99}{990} = \frac{5742}{990} .

5.8 \bar{67} = \frac{5742}{990} + \frac{67}{990} = \frac{5809}{990} .

Step 6: Re-apply the negative sign.

- 5.8 \bar{67} = - \frac{5809}{990} .

Sanity check (decimals): $- 5809 ∕ 990 \approx - 5.867676 \dots$ ✓

ANSWER

$- \frac{5809}{990}$

WATCH OUT

Three places to slip:

Finding $a_{1}$ . It’s the value of the first repeated piece in its actual decimal location, not just the digits “ $67$ .” Here that first piece is $0.067 = 67 ∕ 1000$ , not $67$ on its own.
Finding $r$ . It’s the multiplier that takes you from one block to the next. A block of length $k$ shifts you $k$ decimal places to the right each time, so $r = 1 0^{- k}$ . Here the block “ $67$ ” has length $2$ , so $r = 1 ∕ 100$ .
Handling the non-repeating prefix. Don’t try to bake $5.8$ into the geometric series. Just compute the repeating tail as a series, then add the non-repeating part at the end. (Common denominator $990$ makes the final fraction clean.)

CONNECTION

Faster alternative (the shift-and-subtract trick). Once you trust the geometric-series result, here’s a quicker mechanical method for future repeating-decimal problems. Set $x = - 5.8 \bar{67}$ , then:

10 x = - 58 . \bar{67}, 1000 x = - 5867 . \bar{67} .

Subtracting: $1000 x - 10 x = - 5867 - (- 58) = - 5809$ , so $990 x = - 5809$ and $x = - \frac{5809}{990}$ .

The two methods give the same answer because they’re the same math: $990 = 1 0^{j + k} - 1 0^{j}$ where $j$ is the number of non-repeating decimal digits between the decimal point and the repeating block (here $j = 1$ , the “ $8$ ”) and $k$ is the block length. The shift-and-subtract trick is just a streamlined way to compute the geometric sum without writing the series.

On the test: since the wording says “use a geometric series,” show the series setup explicitly — $a_{1} = 67 ∕ 1000$ , $r = 1 ∕ 100$ , $S = a_{1} ∕ (1 - r) = 67 ∕ 990$ — and then combine with the non-repeating part. Don’t lead with shift-and-subtract on this one.

ABOUT THIS EXCERPT

The guide continues for another twenty-seven pages — every remaining problem worked the same way, through the final Taylor-series questions.

Every family receives the complete guide, after every session.

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Source and review notes

What this resource is

Complete browser edition of a sanitized real-session solution-guide excerpt. Names, dates, and student work were removed; unlike the AP packet series, this page does not make an all-original-problem claim.

Published: July 15, 2026
Last reviewed: August 7, 2026
Printable edition: 8 pages

Read the complete packet above in the browser, or use the PDF for its fixed-page print layout and offline reading.

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Sequences and series

Diagnostic Decision Tree

Master Toolbox — All the Formulas You Need

The 21 Problems — In Review-Packet Order

PROBLEM 1

What this resource is

The next step is making the method fit the student.

Sequences and series

Diagnostic Decision Tree

Master Toolbox — All the Formulas You Need

The 21 Problems — In Review-Packet Order

PROBLEM 1

What this resource is

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