Geometric series — convergence condition, proof: Question bank

This is a conceptual workout for Geometric series — convergence condition & proof. No heavy arithmetic here — every item hunts a misconception or a boundary case. Read each prompt, commit to an answer out loud, THEN reveal. The answer side always gives the reason, not just a verdict.

Before you start, keep three anchors from the parent note in mind:

The series is $\sum a r^{n} = a + a r + a r^{2} + \dots$ , where $== a ==$ is the actual first term you write down and $== r ==$ is the common ratio (each term ÷ previous term).
The closed form $\frac{a}{1 - r}$ is only valid when $== ∣ r ∣ < 1 ==$ , because its derivation secretly used $r^{N} \to 0$ .
"Converges" means the sequence of partial sums $S_{N}$ (see Partial sums and series convergence) approaches a single finite number as $N \to \infty$ .

True or false — justify

True or false: If the terms of a series get smaller and smaller, the series must converge.

False. Shrinking terms are necessary but not sufficient — the harmonic series

1 + \frac{1}{2} + \frac{1}{3} + \dots

has terms

\to 0

yet diverges. For geometric series the shrinking must be fast (constant ratio

∣ r ∣ < 1

), which is stronger than merely

\to 0

True or false:

\sum_{n = 0}^{\infty} a r^{n}

converges if and only if

- 1 < r < 1

True. The condition

∣ r ∣ < 1

unpacks to exactly

- 1 < r < 1

, both endpoints excluded (

r = 1

gives

S_{N} = a N \to \infty

;

r = - 1

makes

r^{N}

oscillate

1, - 1, 1, \dots

and never settle).

True or false: A geometric series with negative

r

always sums to a negative number.

False. With

∣ r ∣ < 1

the sum is

\frac{a}{1 - r}

; a negative

r

makes

1 - r > 1

, so the sum shrinks below

a

but keeps

a

's sign. E.g.

\sum 3 (- \frac{1}{3})^{n} = \frac{9}{4} > 0

True or false: If

∣ r ∣ < 1

the terms must all be positive.

False. Sign of

r

is independent of convergence. With

r = - \frac{1}{2}

the terms alternate sign, yet

∣ r ∣ < 1

so it still converges — the alternation actually helps by partial cancellation.

True or false: The formula

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

holds for every real

r

False. It holds for every

r

except

r = 1

, where the division by

1 - r = 0

is illegal; there

S_{N} = a N

directly. (The formula is a finite sum, so it needs no convergence assumption — only

r \neq = 1

True or false: Doubling the first term

a

doubles the infinite sum.

True. The sum

\frac{a}{1 - r}

is linear in

a

;

r

(which controls convergence) is untouched, so the total scales by the same factor. This is why

a

is a pure "volume knob."

True or false: If a geometric series diverges, its partial-sum formula

\frac{a ( 1 - r ^{N} )}{1 - r}

is wrong.

False. The partial-sum formula stays correct for any finite

N

(as long as

r \neq = 1

). What fails is taking the limit:

r^{N}

doesn't go to

0

, so the limit of

S_{N}

doesn't exist. The formula isn't wrong; the limit step is unavailable.

Spot the error

Find the flaw: "For

\sum_{n = 0}^{\infty} 2 \cdot 3^{n}

, we get

\frac{a}{1 - r} = \frac{2}{1 - 3} = - 1

, so the sum is

- 1

Here

∣ r ∣ = 3 \geq 1

, so the series diverges — the formula is inapplicable. A sum of positive growing terms cannot be a negative number; the "

- 1

" is a meaningless output of an assumption (

r^{N} \to 0

) that is false.

Find the flaw: "For

\sum_{n = 1}^{\infty} (\frac{1}{2})^{n}

, the coefficient is

1

, so

a = 1

and

S = \frac{1}{1 - 1/2} = 2

The first actual term (plug

n = 1

) is

(\frac{1}{2})^{1} = \frac{1}{2}

, so

a = \frac{1}{2}

, giving

S = \frac{1/2}{1/2} = 1

. Grabbing the coefficient out front instead of computing the first written term is the classic

a

-misread.

Find the flaw: "Since

\sum_{n = 0}^{\infty} a r^{n}

and

\sum_{n = 1}^{\infty} a r^{n}

have the same

a

and

r

, they have the same sum."

They differ by exactly the

n = 0

term, which is

a

. So

\sum_{n = 1}^{\infty} a r^{n} = \frac{a}{1 - r} - a = \frac{a r}{1 - r}

. The index start changes the first term, hence the sum.

Find the flaw: "The series

1 - 1 + 1 - 1 + \dots

has

a = 1, r = - 1

, so its sum is

\frac{1}{1 - ( - 1 )} = \frac{1}{2}

With

r = - 1

we have

∣ r ∣ = 1

, not

< 1

, so the series diverges: its partial sums oscillate

1, 0, 1, 0, \dots

and never approach

\frac{1}{2}

. The value

\frac{1}{2}

is a formal fiction (Cesàro/Abel average), not the sum.

Find the flaw: "As

r \to 1^{-}

, the sum

\frac{a}{1 - r}

approaches a large finite number."

r \to 1^{-}

, the denominator

1 - r \to 0^{+}

, so

\frac{a}{1 - r} \to + \infty

(for

a > 0

) — there is no finite limit. The terms barely shrink, so the pile grows without bound.

Find the flaw: "The telescoping trick

S_{N} - r S_{N}

needs

∣ r ∣ < 1

to work."

No — the cancellation is pure algebra on a finite sum and works for any

r \neq = 1

. The condition

∣ r ∣ < 1

is only needed later, when we let

N \to \infty

and require

r^{N} \to 0

Find the flaw: "

\sum_{n = 0}^{\infty} 0 \cdot r^{n}

diverges when

r = 5

because

∣ r ∣ \geq 1

With

a = 0

every term is

0

, so

S_{N} = 0

for all

N

and the sum is

0

— it converges trivially. The "diverges when

∣ r ∣ \geq 1

" rule carries the fine print

a \neq = 0

Why questions

Why does the derivation of

\frac{a}{1 - r}

break down exactly at

∣ r ∣ = 1

Because the closed form comes from

lim_{N \to \infty} (1 - r^{N})

, which needs

r^{N} \to 0

. At

r = 1

r^{N} = 1

forever (no shrink); at

r = - 1

r^{N}

oscillates. Only

∣ r ∣ < 1

forces

r^{N} \to 0

(see Sequences — limits and convergence).

Why do we sum the first

N

terms and take a limit, instead of adding infinitely many terms directly?

You cannot perform infinitely many additions. "Infinite sum" is defined as

lim_{N \to \infty} S_{N}

— the value the finite partial sums approach. This is the whole meaning of series convergence.

Why does a negative ratio make the sum smaller than the first term

a

(when

a > 0

Negative

r

makes terms alternate in sign, so they partially cancel. Algebraically

1 - r > 1

, so

\frac{a}{1 - r} < a

— the bigger denominator directly reflects that cancellation.

Why can't we conclude convergence just from

r^{N} \to 0

term-by-term without the partial-sum formula?

Individual terms

\to 0

never guarantees the sum converges (harmonic series is the counterexample). We need the explicit

S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r}

to see the whole partial sum approach a finite limit.

Why is the geometric series called "the model" for power series?

Setting

a = 1, r = x

gives

\sum x^{n} = \frac{1}{1 - x}

for

∣ x ∣ < 1

. The convergence boundary

∣ x ∣ < 1

becomes the radius of convergence; see Power series and radius of convergence.

Why does the telescoping cancellation leave only two surviving terms?

S_{N} - r S_{N}

, every internal

a r^{k}

appears once in

S_{N}

and once in

r S_{N}

, cancelling. Only the very first term of

S_{N}

(

a

) and the very last of

r S_{N}

(

a r^{N}

) have no partner — the same idea as a Telescoping series.

Why does the Ratio Test give the same threshold

∣ r ∣ < 1

for a geometric series?

The ratio of consecutive terms is exactly

r

(constant), so the Ratio Test's limit

L = ∣ r ∣

; it declares convergence for

L < 1

— identical boundary. Geometric series is the special case where the ratio never varies.

Edge cases

Edge case: What is the sum when

r = 0

Every term after the first vanishes (

0^{n} = 0

for

n \geq 1

), so the series is just

a

. Formula agrees:

\frac{a}{1 - 0} = a

. (By convention

0^{0} = 1

for the

n = 0

term.)

Edge case: What if

a = 0

and

∣ r ∣ \geq 1

, say

r = 10

The series is

0 + 0 + 0 + \dots = 0

, converging to

0

. The divergence rule "

∣ r ∣ \geq 1 \Rightarrow

diverges" assumes

a \neq = 0

; a zero first term kills every term regardless of

r

Edge case: Does

\sum_{n = 0}^{\infty} r^{n}

converge at the endpoint

r = 1

No. It becomes

1 + 1 + 1 + \dots

, so

S_{N} = N \to \infty

. Both endpoints of

(- 1, 1)

fail, so the interval of convergence is open.

Edge case: For

\sum 7 (\frac{1}{10})^{n}

starting at

n = 1

, does the boundary case

∣ r ∣ = 1

ever arise?

No — here

r = \frac{1}{10}

, comfortably inside

∣ r ∣ < 1

, so it converges to

\frac{7/10}{9/10} = \frac{7}{9}

. This is why every repeating decimal is a fraction (see Repeating decimals as fractions); its ratio is always

1 0^{- k}

with

∣ r ∣ < 1

Edge case: Can a geometric series converge to a value with the opposite sign of

a

No. The sum

\frac{a}{1 - r}

has

1 - r > 0

whenever

∣ r ∣ < 1

, so the sign of the sum always matches the sign of

a

. Any "negative sum of positive terms" signals you ignored the

∣ r ∣ < 1

check.

Edge case: What happens to

S_{N}

r = - 1

N

grows?

With

a = 1

, partial sums cycle

1, 0, 1, 0, \dots

— they are bounded but never settle on one value, so the series diverges by oscillation. Boundedness alone is not convergence.

Recall One-line litmus test for every trap above

Before writing $\frac{a}{1 - r}$ : (1) identify the true first term $a$ by plugging in the start index, (2) confirm $== ∣ r ∣ < 1 ==$ , (3) only then apply the formula. Skip step 2 and you'll "prove" nonsense like $2 + 6 + 18 + \dots = - 1$ .

Connections

Parent: convergence condition & proof
Sequences — limits and convergence — the $lim r^{N}$ every trap leans on
Partial sums and series convergence — "converges" defined via $S_{N}$
Ratio Test — same $∣ r ∣ < 1$ threshold, generalised
Power series and radius of convergence — geometric series as the model $\sum x^{n} = \frac{1}{1 - x}$
Repeating decimals as fractions — edge case with $r = 1 0^{- k}$
Telescoping series — the cancellation behind the proof