Infinite Series

♾️ Infinite Series

Part 1 of 5 — From Sequences to Series

Topics in This Part

🔑 Key Concept: A series is what you get when you add up the terms of a sequence. An infinite series keeps adding forever — and the surprising idea of this lesson is that an endless sum can still land on a single, finite number.

Sequence vs. Series

A sequence is an ordered list of numbers. A series is the sum of those numbers.

The numbers themselves are called terms. We label them $a_1, a_2, a_3, \dots$, where $a_1$ is the first term.

Finite vs. Infinite

• A finite series stops: $2 + 4 + 8 + 16$ (four terms).
• An infinite series never stops: $2 + 4 + 8 + 16 + \cdots$ — the $\cdots$ means "forever."

💡 The difference is just punctuation, but it changes everything. A finite sum is always a plain number. An infinite sum might be a number — or it might blow up to infinity.

Concept Check 🎯

Partial Sums

We can't write down an infinite sum all at once, so we sneak up on it using partial sums. The $n$-th partial sum, written $S_n$, adds just the first $n$ terms.

For the series $1 + 2 + 4 + 8 + \cdots$:

$S_1 = 1$ $S_2 = 1 + 2 = 3$ $S_3 = 1 + 2 + 4 = 7$ $S_4 = 1 + 2 + 4 + 8 = 15$

Each partial sum builds on the last: $S_n = S_{n-1} + a_n$.

🔑 Big Idea: To understand an infinite series, we watch what the sequence of partial sums $S_1, S_2, S_3, \dots$ does. If they home in on one number, the infinite series equals that number. (We make this precise in Part 3.)

Compute Partial Sums 🧮

For the series $3 + 6 + 9 + 12 + 15 + \cdots$, find each partial sum.

1) $S_2 = \,?$ 2) $S_3 = \,?$ 3) $S_4 = \,?$

Summation (Sigma) Notation

Writing out long sums is tedious, so mathematicians use the Greek capital sigma $\sum$ as shorthand for "add these up."

$\sum_{n=1}^{4} 2n \;=\; 2(1) + 2(2) + 2(3) + 2(4) \;=\; 2 + 4 + 6 + 8 \;=\; 20$

Read it like a recipe:

• $n = 1$ at the bottom is the starting index.
• $4$ on top is the last index (an $\infty$ here means an infinite series).
• $2n$ is the rule for each term — plug in $n = 1, 2, 3, 4$.

💡 The index letter ($n$, $k$, $i$, …) is just a placeholder — it never appears in the answer.

Concept Check 🎯

♾️ Infinite Series

Part 2 of 5 — Geometric Series & the Common Ratio

🔑 The Star of the Show: The one infinite series you can fully tame in Algebra 2 is the geometric series, where each term is the previous term times a fixed number $r$. Everything in this lesson hinges on $r$.

What Makes a Series Geometric?

A series is geometric when you multiply by the same common ratio $r$ to get from each term to the next:

$a_1 \;+\; a_1 r \;+\; a_1 r^2 \;+\; a_1 r^3 \;+\; \cdots$

To find $r$, divide any term by the one before it:

$r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \cdots$

Example: $3 + 6 + 12 + 24 + \cdots$

$r = \frac{6}{3} = 2, \qquad \frac{12}{6} = 2, \qquad \frac{24}{12} = 2 \;\checkmark$

The ratio is consistent, so the series is geometric with $a_1 = 3$ and $r = 2$.

⚠️ Don't confuse geometric with arithmetic. Arithmetic series add a constant ($2 + 5 + 8 + \cdots$, common difference $3$). Geometric series multiply by a constant. Only geometric series have the convergence story we're after.

Find the Common Ratio 🧮

Find $r$ by dividing a term by the one before it. (Fractions like $1/2$ are fine.)

1) $5 + 15 + 45 + 135 + \cdots,\quad r = \,?$ 2) $16 + 8 + 4 + 2 + \cdots,\quad r = \,?$ 3) $2 - 6 + 18 - 54 + \cdots,\quad r = \,?$

Any Term From the First

Because each step multiplies by $r$, you can jump straight to the $n$-th term:

$a_n = a_1 \cdot r^{\,n-1}$

The exponent is $n - 1$ (not $n$) because the first term has been multiplied by $r$ zero times.

Example: $4 + 12 + 36 + \cdots$ ($a_1 = 4$, $r = 3$)

$a_5 = 4 \cdot 3^{\,5-1} = 4 \cdot 3^4 = 4 \cdot 81 = 324$

💡 Sanity check the exponent: for the first term, $a_1 = a_1 \cdot r^{\,1-1} = a_1 \cdot r^0 = a_1 \cdot 1 = a_1$. ✓

Classify Each Series 🔽

For each series, choose its common ratio $r$.

Putting $a_1$ and $r$ to Work

Notice that once you know the first term $a_1$ and the ratio $r$, the entire series is locked in — every term, and (as we'll see) the whole sum. The $n$-th-term formula is your tool for reaching any single term quickly.

$a_n = a_1 \cdot r^{\,n-1}$

Let's drill it once more before moving on to the convergence question.

Use $a_n = a_1 r^{\,n-1}$ 🧮

1) For $2 + 6 + 18 + \cdots$, find $a_4$. 2) For $80 + 40 + 20 + \cdots$, find $a_5$.

♾️ Infinite Series

Part 3 of 5 — Convergence: When Does an Infinite Sum Have a Value?

🔑 The Whole Question: Add infinitely many positive numbers and you might expect the total to be infinite. Sometimes it is. But if the terms shrink fast enough, the partial sums settle on a finite number. That settling-down is called convergence.

Watch the Partial Sums

Take the famous series $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \cdots$ ($a_1 = \tfrac12$, $r = \tfrac12$).

The partial sums creep toward $1$ but never overshoot it. We say the series converges to $1$.

Now compare $1 + 2 + 4 + 8 + \cdots$ ($r = 2$): the partial sums $1, 3, 7, 15, 31, \dots$ explode. This series diverges — it has no finite sum.

💡 The difference is the size of $r$. When $|r| < 1$ the terms shrink toward $0$ and the sum settles. When $|r| \ge 1$ the terms don't shrink, so the sum runs away.

The Convergence Rule

🔑 The Rule: An infinite geometric series converges if and only if $\;|r| < 1\;$ (that is, $-1 < r < 1$).

• If $|r| < 1$ → converges (has a finite sum).
• If $|r| \ge 1$ → diverges (no finite sum).

⚠️ Watch the absolute value. A ratio of $r = -\tfrac{1}{2}$ converges because $|-\tfrac12| = \tfrac12 < 1$. Don't reject a series just because $r$ is negative — check the size of $r$, ignoring its sign.

Concept Check 🎯

A Word on Negative Ratios

The trickiest cases are negative ratios, because the signs flip back and forth. The convergence test is exactly the same — just look at $|r|$:

💡 The last row is the boundary case $r = -1$. Its partial sums bounce $1, 0, 1, 0, \dots$ forever and never settle on one value, so it diverges. The boundary $|r| = 1$ is always divergent.

Converge or Diverge? 🔽

Decide whether each infinite geometric series has a finite sum.

Exactly Where Is the Line?

The convergence condition $-1 < r < 1$ is a strict double inequality. Both endpoints are excluded:

• At $r = 1$, every term equals $a_1$, so the sum grows without bound ($a_1 + a_1 + a_1 + \cdots$).
• At $r = -1$, the partial sums oscillate and never settle.

⚠️ "$|r| < 1$" means strictly less than $1$ — the equals case is out. Let's pin down the boundary numerically.

The Boundary 🧮

The convergence rule is $-1 < r < 1$.

1) What is the largest integer value of $r$ for which a geometric series still converges? (Hint: $r$ must be strictly less than $1$.) 2) Does $r = -1$ give convergence? Enter 1 for yes or 0 for no.

♾️ Infinite Series

Part 4 of 5 — The Sum Formula & Repeating Decimals

🔑 The Payoff: When a geometric series converges, we don't just know it has a sum — we can compute that sum exactly with one clean formula.

The Infinite Geometric Sum Formula

For an infinite geometric series with first term $a_1$ and ratio $|r| < 1$:

$S = \frac{a_1}{1 - r}$

That's it. Plug in the first term and the ratio.

Example: $\tfrac12 + \tfrac14 + \tfrac18 + \cdots$ ($a_1 = \tfrac12$, $r = \tfrac12$)

$S = \frac{\tfrac12}{1 - \tfrac12} = \frac{\tfrac12}{\tfrac12} = 1$

This matches the partial sums creeping toward $1$ from Part 3. ✓

Example: $6 + 2 + \tfrac23 + \cdots$ ($a_1 = 6$, $r = \tfrac13$)

$S = \frac{6}{1 - \tfrac13} = \frac{6}{\tfrac23} = 6 \cdot \frac{3}{2} = 9$

⚠️ Check $|r| < 1$ first! The formula only works for convergent series. If $|r| \ge 1$, there is no sum, and plugging into $\frac{a_1}{1-r}$ produces a meaningless number. Always verify convergence before you compute.

Apply the Formula 🧮

Use $S = \dfrac{a_1}{1-r}$. Each series converges, so go ahead.

1) $8 + 4 + 2 + 1 + \cdots\quad(a_1 = 8,\; r = \tfrac12)$ 2) $3 + 1 + \tfrac13 + \cdots\quad(a_1 = 3,\; r = \tfrac13)$ 3) $10 - 5 + 2.5 - \cdots\quad(a_1 = 10,\; r = -\tfrac12)$

A Surprising Use: Repeating Decimals Are Fractions

A repeating decimal is secretly an infinite geometric series. This is why every repeating decimal equals a fraction.

Example: $0.\overline{3} = 0.3333\ldots$

Break it into place values:

$0.\overline{3} = 0.3 + 0.03 + 0.003 + \cdots = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$

This is geometric with $a_1 = \tfrac{3}{10}$ and $r = \tfrac{1}{10}$:

$S = \frac{\tfrac{3}{10}}{1 - \tfrac{1}{10}} = \frac{\tfrac{3}{10}}{\tfrac{9}{10}} = \frac{3}{9} = \frac{1}{3}$

So $0.\overline{3} = \tfrac13$ — exactly. ✓

Example: $0.\overline{27} = 0.272727\ldots$

Group in pairs: $a_1 = \tfrac{27}{100}$, $r = \tfrac{1}{100}$:

$S = \frac{\tfrac{27}{100}}{1 - \tfrac{1}{100}} = \frac{\tfrac{27}{100}}{\tfrac{99}{100}} = \frac{27}{99} = \frac{3}{11}$

💡 The repeating block sets both $a_1$ and $r$: a $1$-digit block gives $r = \tfrac1{10}$, a $2$-digit block gives $r = \tfrac1{100}$, a $3$-digit block gives $r = \tfrac1{1000}$, and so on.

Concept Check 🎯

The Two-Step Recipe

Every "repeating decimal → fraction" problem follows the same two steps:

1. Set up $a_1$ and $r$ from the repeating block. The block becomes the numerator of $a_1$; the number of digits $k$ in the block makes $r = \dfrac{1}{10^k}$.
2. Plug into $S = \dfrac{a_1}{1-r}$ and reduce the fraction.

💡 A handy shortcut falls out of this: $0.\overline{d_1 d_2 \cdots d_k} = \dfrac{(\text{the } k\text{-digit block})}{\underbrace{99\cdots9}_{k\text{ nines}}}$. For example, $0.\overline{27} = \tfrac{27}{99}$ and $0.\overline{6} = \tfrac{6}{9} = \tfrac23$. Your turn:

Repeating Decimals → Fractions 🧮

Convert each repeating decimal to a fraction using $S = \dfrac{a_1}{1-r}$. Enter your answer as a fraction in lowest terms (e.g. 3/11).

1) $0.\overline{1} \;(a_1 = \tfrac1{10},\; r = \tfrac1{10})$ 2) $0.\overline{45} \;(a_1 = \tfrac{45}{100},\; r = \tfrac1{100})$

♾️ Infinite Series

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) read summation notation, (2) find a common ratio, (3) decide convergence with the $|r| < 1$ rule, and (4) sum a convergent series with $S = \dfrac{a_1}{1-r}$. Let's put it all together.

Quick Reference

⚠️ The #1 trap: Using $S = \frac{a_1}{1-r}$ on a divergent series. Always confirm $|r| < 1$ before you compute a sum — if $|r| \ge 1$, the correct answer is "diverges (no sum)."

Mixed Practice 🎯

A Reliable Game Plan

Whenever you meet an infinite geometric series, run this checklist:

1. Find $r$ by dividing a term by the one before it.
2. Check $|r| < 1$. If not, stop and answer "diverges."
3. Identify $a_1$ (the first term).
4. Compute $S = \dfrac{a_1}{1 - r}$ and simplify.

🔑 Steps 1–2 are non-negotiable. Most wrong answers come from skipping the convergence check and blindly applying the formula. Walk through the plan one piece at a time below.

Build the Solution 🔽

You're summing the infinite series $\;\dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + \cdots$. Choose what goes in each step.

You're Ready

You've covered the full arc: series and partial sums, summation notation, common ratios, the $|r| < 1$ convergence test, the sum formula $S = \dfrac{a_1}{1-r}$, and the slick repeating-decimal application.

💡 One last reminder before the quiz: an infinite sum can be finite — but only when the terms shrink fast enough, i.e. when $|r| < 1$. Keep that test front and center.

Finish strong with the Exit Quiz below. ✅

Exit Quiz ✅

Answer all three to finish the lesson.