Chapter 09

Medium

Sequences and Series

00 · Symbol Glossary

$\{a_n\}$a sub n — sequence

An ordered, infinite list of numbers $a_1, a_2, a_3, \ldots$ , indexed by the positive integers $n=1,2,3,\ldots$ . Read "the sequence a-sub-n." Distinct from a function $f(x)$ (Chapter 01): a sequence is only defined at whole-number inputs, not on a continuum.

$\displaystyle\lim_{n\to\infty} a_n$limit of a sequence

The value the terms $a_n$ approach as $n$ grows through the positive integers without bound. Defined the same way as the limits-at-infinity of Chapter 01, Section 04, but with $n$ restricted to integers instead of $x$ ranging continuously.

$\displaystyle\sum_{n=1}^{\infty} a_n$infinite series

The sum of every term of a sequence $\{a_n\}$ , added one after another forever. Read "the sum from $n$ equals $1$ to infinity of $a_n$ ." Whether this infinite sum has any meaning at all is the central question of this chapter.

$S_N$S sub N — partial sum

The finite sum of just the first $N$ terms of a series: $S_N = a_1+a_2+\cdots+a_N$ . A single, ordinary, finite number — no limit involved yet.

$r$r — common ratio

In a geometric sequence, the fixed number each term is multiplied by to get the next: $a_{n+1} = r\cdot a_n$ .

01 · Sequences and Their Limits

A sequence is simply an infinite, ordered list of numbers: $2, 4, 6, 8, \ldots$ or $1, \tfrac12, \tfrac13, \tfrac14, \ldots$ . Each entry is labeled by its position $n=1,2,3,\ldots$ , and the whole list is written $\{a_n\}$ .

Definition — Sequence and Sequence Limit

A sequence $\{a_n\}$ assigns a real number $a_n$ to each positive integer $n$ .

The sequence converges to $L$ if $\displaystyle\lim_{n\to\infty} a_n = L$ — for every $\varepsilon>0$ (Chapter 01's tolerance), there is some integer $N$ such that $\lvert a_n - L\rvert < \varepsilon$ for all $n>N$ . If no such $L$ exists, the sequence diverges.

Example — A converging sequence

$a_n = \dfrac{1}{n}$ : the terms are $1, \tfrac12, \tfrac13, \tfrac14, \ldots$ , clearly shrinking toward $0$ . Formally, $\displaystyle\lim_{n\to\infty}\frac1n = 0$ — the same limit-at-infinity computation as Chapter 01, Section 04, just restricted to integer inputs. The sequence converges to $0$ .

❌ Assuming every sequence with a pattern converges

Consider $a_n = (-1)^n$ : the terms alternate $-1, 1, -1, 1, \ldots$ forever, following a perfectly clear, predictable pattern.

Why it breaks: convergence requires the terms to settle near a single value $L$ as $n$ grows. Here the terms never settle — they permanently oscillate between two fixed values, $1$ and $-1$ , no matter how large $n$ gets. For any proposed $L$ , infinitely many terms stay at distance at least $1$ from $L$ (whichever of $-1,1$ is farther from $L$ ), so no $\varepsilon<1$ can ever be satisfied for all sufficiently large $n$ .

Consequence: $a_n=(-1)^n$ diverges — having a clear, predictable pattern is not the same as converging to a limit.

02 · Series and Partial Sums

Adding up all the terms of a sequence forever produces a series. Since an infinite sum cannot be computed by direct addition, its meaning has to be built from something already understood: ordinary, finite sums.

Definition — Series as a Limit of Partial Sums

Given a sequence $\{a_n\}$ , form the partial sums $S_N = \displaystyle\sum_{n=1}^{N} a_n = a_1+a_2+\cdots+a_N$ — each one an ordinary finite sum.

The series $\displaystyle\sum_{n=1}^{\infty} a_n$ converges to $S$ if $\displaystyle\lim_{N\to\infty} S_N = S$ (an ordinary sequence limit, Section 01). If that limit does not exist or is infinite, the series diverges.

A series is a sequence in disguise

This definition converts the unfamiliar question "does an infinite sum make sense?" into the familiar one from Section 01: "does the sequence of partial sums $\{S_N\}$ converge?" Every series-convergence question in this chapter and Chapter 12 is secretly a sequence-limit question about $\{S_N\}$ .

03 · Geometric Series

Definition — Geometric Series

A geometric series has the form $\displaystyle\sum_{n=0}^{\infty} a\,r^n = a + ar + ar^2 + ar^3 + \cdots$ , where $a\neq0$ is the first term and $r$ is the common ratio.

\sum_{n=0}^{\infty} a\,r^n = \frac{a}{1-r} \quad \text{if } \lvert r\rvert < 1, \qquad \text{diverges if } \lvert r\rvert \geq 1

Step-by-step — Derive the partial sum formula and evaluate $\displaystyle\sum_{n=0}^{\infty} 3\left(\frac12\right)^n$

Write the partial sum:

S_N = a + ar + ar^2 + \cdots + ar^{N-1}

(the first

N

terms, indices

n=0

N-1

Multiply the whole sum by $r$ :

rS_N = ar + ar^2 + \cdots + ar^N

— every term of

S_N

shifted up one power.

Subtract, and watch the middle terms cancel:

S_N - rS_N = a - ar^N

(every term except the first of

S_N

and the last of

rS_N

cancels in pairs).

Factor and solve for $S_N$ :

S_N(1-r) = a(1-r^N) \implies S_N = \dfrac{a(1-r^N)}{1-r}

, valid for

r\neq1

Take the limit as $N\to\infty$ , using $\lvert r\rvert<1 \implies r^N\to0$ :

\displaystyle\lim_{N\to\infty} S_N = \frac{a(1-0)}{1-r} = \frac{a}{1-r}

Apply to $a=3$ , $r=\frac12$ : since

\lvert\frac12\rvert<1

, the series converges:

\dfrac{3}{1-\frac12} = \dfrac{3}{\frac12} = 6

❌ Applying the geometric series formula when |r| ≥ 1

For $\displaystyle\sum_{n=0}^{\infty} 2^n = 1+2+4+8+\cdots$ , plugging $a=1$ , $r=2$ into $\dfrac{a}{1-r}$ gives $\dfrac{1}{1-2} = -1$ .

Why it breaks: the formula's derivation (step 5 above) explicitly used $\lvert r\rvert<1$ to conclude $r^N\to0$ . Here $r=2>1$ , so $r^N=2^N\to\infty$ instead, and step 5's limit does not exist as a finite number — the partial sums $S_N=1+2+4+\cdots+2^{N-1}=2^N-1$ grow without bound.

Consequence: the series diverges to $+\infty$ ; the formula's output of $-1$ is not the sum of a divergent series of entirely positive terms — a nonsensical negative value is a clear signal the convergence hypothesis was violated. The formula is only valid for $\lvert r\rvert<1$ , never outside that range.

Geometric intuition

Each partial sum $S_N$ adds a shrinking correction to the previous one when $\lvert r\rvert<1$ — the terms $ar^n$ shrink geometrically toward $0$ , so the running total settles down instead of growing forever, converging to a finite ceiling $\frac{a}{1-r}$ . When $\lvert r\rvert\geq1$ , each new term is at least as large as the last, so the running total never settles.

04 · The Test for Divergence

Before attempting to find the exact sum of a series (often impossible), a fast preliminary check can rule out convergence entirely.

Definition — Test for Divergence (nth-Term Test)

If $\displaystyle\lim_{n\to\infty} a_n \neq 0$ (including the case where the limit does not exist), then $\displaystyle\sum_{n=1}^{\infty} a_n$ diverges.

Equivalently: if a series converges, its terms must shrink to $0$ . This test can only ever prove divergence — it can never prove convergence, even when $a_n\to0$ (Section 05 shows exactly this).

Example — Quick divergence check

$\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n+1}$ : the terms $a_n=\dfrac{n}{n+1} \to 1 \neq 0$ as $n\to\infty$ (dividing numerator and denominator by $n$ : $\dfrac{1}{1+1/n}\to\dfrac{1}{1+0}=1$ ). Since the terms do not shrink to $0$ , the series diverges by the test for divergence — no partial-sum computation needed.

05 · The Harmonic Series — Terms Shrink to Zero, Yet the Series Diverges

❌ Terms → 0 does NOT guarantee convergence

Consider the harmonic series $\displaystyle\sum_{n=1}^{\infty}\frac1n = 1+\frac12+\frac13+\frac14+\cdots$ . Here $a_n=\dfrac1n\to0$ as $n\to\infty$ (Section 01's example) — the test for divergence gives no information, since it only detects divergence when terms fail to vanish.

Why intuition breaks down: group the terms as $1 + \left(\frac12\right) + \left(\frac13+\frac14\right) + \left(\frac15+\frac16+\frac17+\frac18\right) + \cdots$ . Each parenthesized group of $2^k$ consecutive terms starting at $\frac{1}{2^k+1}$ sums to more than $\frac12$ : for instance $\frac13+\frac14 > \frac14+\frac14=\frac12$ , and $\frac15+\frac16+\frac17+\frac18 > 4\times\frac18=\frac12$ (replacing every term with the smallest one in the group, which only makes the sum smaller, yet it's still bigger than $\frac12$ ).

Consequence: the partial sums pick up more than $\frac12$ infinitely many times, so $S_N\to\infty$ — the harmonic series diverges, despite its terms individually shrinking all the way to $0$ . This is the single most important cautionary example in the whole topic of series: shrinking terms are necessary for convergence (Section 04) but never sufficient.

06 · Exercises

EXERCISE 9.1

Divide numerator and denominator by the highest power of $n$ present, then take the limit term by term.

$a_n = \dfrac{3n^2+1}{n^2+5n}$ . Divide by $n^2$ : $\dfrac{3+1/n^2}{1+5/n}$ . As $n\to\infty$ : $\dfrac{3+0}{1+0}=3$ . The sequence converges to $3$ .

Determine whether the sequence $a_n = \dfrac{3n^2+1}{n^2+5n}$ converges, and if so, to what value.

EXERCISE 9.2

Identify the first term $a$ and common ratio $r$ , check $\lvert r\rvert<1$ , then apply the geometric series formula.

$\displaystyle\sum_{n=0}^{\infty} 5\left(-\frac13\right)^n$ : $a=5$ , $r=-\frac13$ , $\lvert r\rvert=\frac13<1$ , converges. Sum $=\dfrac{5}{1-(-1/3)}=\dfrac{5}{4/3}=\dfrac{15}{4}$ .

Evaluate $\displaystyle\sum_{n=0}^{\infty} 5\left(-\frac13\right)^n$ .

EXERCISE 9.3

Check whether the terms themselves approach $0$ ; if not, the test for divergence settles the question immediately.

$a_n = \dfrac{2n+1}{n}=2+\dfrac1n \to 2\neq0$ . By the test for divergence, $\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n}$ diverges.

Show that $\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n}$ diverges, using the test for divergence.

EXERCISE 9.4

The test for divergence cannot prove convergence — recall which specific series from this chapter has terms shrinking to zero but still diverges.

The test for divergence only detects the case $a_n\not\to0$ ; it says nothing when $a_n\to0$ . The harmonic series $\sum\frac1n$ (Section 05) has $a_n=\frac1n\to0$ yet diverges — so $a_n\to0$ passing the test for divergence's condition is consistent with either convergence or divergence, and a different tool is required to tell them apart.

Explain why "the terms of a series approach $0$ " is not enough, by itself, to conclude the series converges. Name a specific series from this chapter that demonstrates this.

EXERCISE 9.5

Write out the first several partial sums directly and look for a pattern of cancellation, rather than trying to force the geometric series formula.

$\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)$ . Partial sum: $S_N = \left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)$ . Every interior term cancels (a "telescoping" pattern), leaving $S_N = 1-\frac{1}{N+1}$ . As $N\to\infty$ : $S_N\to1-0=1$ . The series converges to $1$ .

Evaluate $\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$ by writing out and simplifying the partial sum $S_N$ directly.

EXERCISE 9.6

A perpetuity-style repeated cash flow with a fixed discount factor each period is exactly a geometric series — identify $a$ and $r$ from the payment schedule.

A payment of $100$ dollars arrives at the end of each year forever, discounted by a factor of $0.95$ per year (i.e. year- $n$ payment is worth $100(0.95)^n$ dollars today, for $n=1,2,3,\ldots$ ). Present value: $\displaystyle\sum_{n=1}^{\infty}100(0.95)^n = 100(0.95)\sum_{n=0}^{\infty}(0.95)^n = 95\cdot\dfrac{1}{1-0.95} = 95\cdot20=1900$ . Total present value: $1900$ dollars.

A payment of $100$ dollars arrives at the end of each year forever; each year's payment is discounted by an additional factor of $0.95$ compared to the year before. Model the total present value as a geometric series and evaluate it.

07 · Chapter Summary

Concept	Formula / Rule
Sequence	$\{a_n\}$ , defined only at integers $n=1,2,3,\ldots$
Sequence convergence	$\lim_{n\to\infty}a_n = L$ (finite)
Partial sum	$S_N = \sum_{n=1}^N a_n$ , an ordinary finite sum
Series convergence	$\sum_{n=1}^\infty a_n$ converges iff $\lim_{N\to\infty}S_N$ exists (finite)
Geometric series	$\sum a r^n = \dfrac{a}{1-r}$ if $\lvert r\rvert<1$ ; diverges otherwise
Test for divergence	$a_n\not\to0 \implies$ series diverges (never proves convergence)
Harmonic series	$\sum\frac1n$ diverges despite $a_n\to0$

Next: Chapter 10 — Power Series and Taylor Series, where a series with variable terms $c_n(x-a)^n$ is used to represent functions themselves.